Parabola - JEE Advanced Previous Year Questions with Solutions
Parabola JEE Advanced previous year questions test concepts like tangents, normals, focal chords, locus problems, and common tangents to circles. From 2009 to 2016, JEE Advanced asked 15+ questions on parabola — predominantly multi-correct and paragraph types — making it one of the highest-yield conic section topics. Mastering these solutions can add 12–16 marks to your score.
JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.Download eSaral app for free study material and video tutorials.
Q. The tangent PT and the normal PN to the parabola $y^{2}$ = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose (A) vertex is $\left(\frac{2 \mathrm{a}}{3}, 0\right)$(B) directrix is x = 0(C) latus rectum is $\frac{2 \mathrm{a}}{3}$(D) focus is (a, 0)[JEE 2009, 4]
Ans. (A,D)
Q. Let A and B be two distinct point on the parabola $y^{2}$ = 4x. If the axis of the parabola touches a circle of radius r having AB as its diameter, then the slope of the line joining A and B can be – (A) $\frac{-1}{\mathrm{r}}$(B) $\frac{1}{\mathrm{r}}$(C) $\frac{2}{\mathrm{r}}$(D) $\frac{-2}{\mathrm{r}}$[JEE 2010,3]
Ans. (C,D) $\mathrm{t}_{1}+\mathrm{t}_{2}=\mathrm{r}$$\frac{2}{\mathrm{r}}=\frac{2}{\mathrm{t}_{1}+\mathrm{t}_{2}}$similarly $-\frac{2}{\mathrm{r}}$ isalso possible
Q. Consider the parabola $\mathrm{y}^{2}=8 \mathrm{x}$. Let $\Delta_{1}$ be the area of the triangle formed by the end points of its latus rectum and the point $\mathrm{P}\left(\frac{1}{2}, 2\right)$ on the parabola, and $\Delta_{2}$ be the area of the triangle formed by drawing tangents at $\mathrm{P}$ and at the end points of the latus rectum. Then $\frac{\Delta_{1}}{\Delta_{2}}$ is [JEE 2011,4]
Ans. 2
Q. Let (x,y) be any point on the parabola $\mathrm{y}^{2}$ = 4x. Let P be the point that divides the line segment from (0,0) to (x,y) in the ratio 1 : 3. Then the locus of P is- (A) $x^{2}=y$(B) $\mathrm{y}^{2}=2 \mathrm{x}$(C) $\mathrm{y}^{2}=\mathrm{x}$(D) $x^{2}=2 y$[JEE 2011,3]
Ans. (C) Let P be (h, k)on using section formula $\mathrm{P}\left(\frac{x}{4}, \frac{y}{4}\right)$$\therefore \quad \mathrm{h}=\frac{x}{4}$ and $\mathrm{k}=\frac{y}{4}$$\Rightarrow \quad \mathrm{x}=4 \mathrm{h}$ and $\mathrm{y}=4 \mathrm{k}$$\because \quad(\mathrm{x}, \mathrm{y})$ lies on $\mathrm{y}^{2}=4 \mathrm{x}$$\therefore \quad 16 \mathrm{k}^{2}=16 \mathrm{h} \quad \Rightarrow \mathrm{k}^{2}=\mathrm{h}$Locus of point $\mathrm{P}$ is $\mathrm{y}^{2}=\mathrm{x}$
Q. Let L be a normal to the parabola $\mathrm{y}^{2}=4 \mathrm{x} .$ If $\mathrm{L}$ passes through the point $(9,6),$ then $\mathrm{L}$ is given by $-$ (A) y – x + 3 =0(B) y + 3x – 33 = 0(C) y + x – 15 = 0(D) y – 2x + 12 = 0[JEE 2011,3]
Ans. (A,B,D) Equation of normal is $\mathrm{y}=\mathrm{mx}-2 \mathrm{m}-\mathrm{m}^{3}$It passes through the point $(9,6)$ then$6=9 \mathrm{m}-2 \mathrm{m}-\mathrm{m}^{3}$$\Rightarrow \mathrm{m}^{3}-7 \mathrm{m}+6=0$$\Rightarrow(\mathrm{m}-1)(\mathrm{m}-2)(\mathrm{m}+3)=0$$\Rightarrow \quad \mathrm{m}=1,2,-3$Equations of normals are$\mathrm{y}-\mathrm{x}+3=0, \mathrm{y}+3 \mathrm{x}-33=0$& $\quad \mathrm{y}-2 \mathrm{x}+12=0$
Q. Let S be the focus of the parabola $y^{2}=8 x \&$ let PQ be the common chord of the circle $x^{2}+y^{2}-2 x-4 y$ $=0$ and the given parabola. The area of the triangle PQS is [JEE 2012, 4M]
Ans. 4
Paragraph for Question 7 and 8Let PQ be a focal chord of the parabolas $\mathrm{y}^{2}=4 \mathrm{ax} .$ The tangents to the parabola at $\mathrm{P}$ and $\mathrm{Q}$ meet at a point lying on the line $\mathrm{y}=2 \mathrm{x}+\mathrm{a}, \mathrm{a}>0 .$
Q. If chord PQ subtends an angle $\theta$ at the vertex of $y^{2}=4 a x,$ then $\tan \theta=$ (A) $\frac{2}{3} \sqrt{7}$(B) $\frac{-2}{3} \sqrt{7}$(C) $\frac{2}{3} \sqrt{5}$(D) $\frac{-2}{3} \sqrt{5}$[JEE(Advanced) 2013, 3, (–1)]
Ans. (D) Single tangent at the extrimities of a focalchord will intersect on directrix.$\therefore \quad \mathrm{M}\left(-\mathrm{a}, \mathrm{a}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)\right)$lies on $\mathrm{y}=2 \mathrm{x}+\mathrm{a}$$\mathrm{a}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)=-2 \mathrm{a}+\mathrm{a} \quad \Rightarrow \quad \mathrm{t}_{1}+\mathrm{t}_{2}=-1$$\quad \& \quad \mathrm{t}_{1} \mathrm{t}_{2}=-1$$\tan \theta=\left(\frac{\frac{2}{t_{1}}-\frac{2}{t_{2}}}{1+\frac{4}{t_{1} t_{2}}}\right)=\left(\frac{2\left(t_{2}-t_{1}\right)}{3}\right)$$\because\left(t_{2}-t_{1}\right)=\left(t_{2}+t_{1}\right)^{2}-4 t_{1} t_{2}=5$$t_{2}-t_{1}=\pm \sqrt{5}$$\therefore \quad \tan \theta=\pm \frac{2 \sqrt{5}}{3}$but $\theta$ is obtuse because $\mathrm{O}$ is the interior point of the circle for which $\mathrm{PQ}$ is diameter. $\therefore \quad \tan \theta=\frac{-2 \sqrt{5}}{3}$
Q. Length of chord PQ is (A) 7a (B) 5a (C) 2a (D) 3a[JEE(Advanced) 2013, 3, (–1)]
Q. A line L: $y=m x+3$ meets $y-$ axis at $E(0,3)$ and the arc of the parabola $y^{2}=16 x, 0 \leq y$ $\leq 6$ at the point $F\left(x_{0}, y_{0}\right)$. The tangent to the parabola at $F\left(x_{0}, y_{0}\right)$ intersects the $y$ -axis at $G\left(0, y_{1}\right) .$The slope $m$ of the line $L$ is chosen such that the area of the triangle EFG has a local maximum.Match List-I with List-II and select the correct answer using the code given below the lists.[JEE(Advanced) 2013, 3, (–1)]
Ans. (A )
Q. The common tangents to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=2$ and the parabola $\mathrm{y}^{2}=8 \mathrm{x}$ touch the circle at the point $\mathrm{P}, \mathrm{Q}$ and the parabola at the points $\mathrm{R}, \mathrm{S}$. Then the area of the quadrilateral PQRS is –(A) 3 (B) 6 (C) 9 (D) 15[JEE(Advanced)-2014, 3(–1)]
Ans. (D)
Paragraph For Questions 11 and 12Let a,r,s, t be nonzero real numbers. Let $P\left(a t^{2}, 2 a t\right), Q, R\left(a r^{2}, 2 a r\right)$ and $S\left(a s^{2}, 2 a s\right)$ be distinct points on the parabola $y^{2}=4 a x .$ Suppose that $P Q$ is the focal chord and lines $Q R$ and $P K$ are parallel, where $K$ is the point $(2 a, 0) .$The value of $r$ is-
Q. The value of r is- (A) $-\frac{1}{\mathfrak{t}}$(B) $\frac{t^{2}+1}{t}$(C) $\frac{1}{\mathrm{t}}$(D) $\frac{\mathfrak{t}^{2}-1}{\mathfrak{t}}$[JEE(Advanced)-2014, 3(–1)]
Ans. (D) $\because \mathrm{PQ}$ is a focal chord$\therefore$ co-ordinates of point $\mathrm{Q}$ are $=\left(\frac{\mathrm{a}}{\mathrm{t}^{2}},-\frac{2 \mathrm{a}}{\mathrm{t}}\right)$$\mathrm{m}_{\mathrm{QR}}=\frac{2 \mathrm{a}\left(\mathrm{r}+\frac{1}{\mathrm{t}}\right)}{\mathrm{a}\left(\mathrm{r}^{2}-\frac{1}{\mathrm{t}^{2}}\right)}=\frac{2}{\left(\mathrm{r}-\frac{1}{\mathrm{t}}\right)}$$\mathrm{m}_{\mathrm{PK}}=\frac{2 \mathrm{at}-0}{\mathrm{a}\left(\mathrm{t}^{2}-2\right)}=\frac{2 \mathrm{t}}{\mathrm{t}^{2}-2}$Given $\mathrm{m}_{\mathrm{QR}}=\mathrm{m}_{\mathrm{PK}}$$\Rightarrow \frac{2}{\mathrm{r}-\frac{1}{\mathrm{t}}}=\frac{2 \mathrm{t}}{\mathrm{t}^{2}-2} \Rightarrow \mathrm{r}=\frac{\mathrm{t}^{2}-2}{\mathrm{t}}+\frac{1}{\mathrm{t}}$$\Rightarrow \mathrm{r}=\mathrm{t}-\frac{2}{\mathrm{t}}+\frac{1}{\mathrm{t}} \Rightarrow \mathrm{r}=\frac{\mathrm{t}^{2}-1}{\mathrm{t}}$
Q. If st = 1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is- (A) $\frac{\left(\mathrm{t}^{2}+1\right)^{2}}{2 \mathrm{t}^{3}}$(B) $\frac{\mathrm{a}\left(\mathrm{t}^{2}+1\right)^{2}}{2 \mathrm{t}^{3}}$(C) $\frac{\mathrm{a}\left(\mathrm{t}^{2}+1\right)^{2}}{\mathrm{t}^{3}}$(D) $\frac{\mathrm{a}\left(\mathrm{t}^{2}+2\right)^{2}}{\mathrm{t}^{3}}$[JEE(Advanced)-2014, 3(–1)]
Ans. (B)
Q. If the normals of the parabola $y^{2}=4 x$ drawn at the end points of its latus rectum are tangents to the circle $(x-3)^{2}+(y+2)^{2}=r^{2},$ then the value of $r^{2}$ is [JEE 2015, 4M, –0M]
Ans. 2
Q. Let the curve $C$ be the mirror image of the parabola $y^{2}=4 x$ with respect to the line $x+y+4=0 .$ If $A$ and $B$ are the points of intersection of $C$ with the line $y=-5$, then the distance between A and $B$ is [JEE 2015, 4M, –0M]
Ans. 4 Let there be a point $\left(\mathrm{t}^{2}, 2 \mathrm{t}\right)$ on $\mathrm{y}^{2}=4 \mathrm{x}$Clearly its reflection in $\mathrm{x}+\mathrm{y}+4=0$ is given by$$\frac{\mathrm{x}-\mathrm{t}^{2}}{1}=\frac{\mathrm{y}-2 \mathrm{t}}{1}=\frac{-2\left(\mathrm{t}^{2}+2 \mathrm{t}+4\right)}{2}$$$\therefore \quad \mathrm{x}=-(2 \mathrm{t}+4) \quad \& \mathrm{y}=-\left(\mathrm{t}^{2}+4\right)$Now, $\mathrm{y}=-5 \quad \Rightarrow \quad \mathrm{t}=\pm 1$$\therefore \quad \mathrm{x}=-6 \quad$ or $\quad \mathrm{x}=-2$$\therefore \quad$ Distance between $\mathrm{A} \& \mathrm{B}=4$
Q. Let $P$ and $Q$ be distinct points on the parabola $y^{2}=2 x$ such that a circle with $P Q$ as diameter passes through the vertex $O$ of the parabola. If $P$ lies in the first quadrant and the area of the triangle $\Delta O P Q$ is $3 \sqrt{2},$ then which of the following is (are) the coordinates of $P ?$ (A) $(4,2 \sqrt{2})$(B) $(9,3 \sqrt{2})$(C) $\left(\frac{1}{4}, \frac{1}{\sqrt{2}}\right)$(D) $(1, \sqrt{2})$[JEE 2015, 4M, –2M]
Ans. (A,D)
Q. The circle $C_{1}: x^{2}+y^{2}=3,$ with centre at $O,$ intersects the parabola $x^{2}=2 y$ at the point $P$ in the first quadrant. Let the tangent to the circle $C_{1}$ at $P$ touches other two circles $C_{2}$ and $C_{3}$ at $R_{2}$ and $R_{3},$ respectively. Suppose $C_{2}$ and $C_{3}$ have equal radii $2 \sqrt{3}$ and centres $Q_{2}$ and $Q_{3}$ respectively. If $Q_{2}$ and $Q_{3}$ lie on the y-axis, then- (A) $\mathrm{Q}_{2} \mathrm{Q}_{3}=12$(B) $\mathrm{R}_{2} \mathrm{R}_{3}=4 \sqrt{6}$(C) area of the triangle $\mathrm{OR}_{2} \mathrm{R}_{3}$ is $6 \sqrt{2}$(D) area of the triangle $\mathrm{PQ}_{2} \mathrm{Q}_{3}$ is $4 \sqrt{2}$[JEE (Advanced) 2016]
Ans. (A,B,C) On solving $\mathrm{x}^{2}+\mathrm{y}^{2}=3$ and $\mathrm{x}^{2}=2 \mathrm{y}$ we get point $\mathrm{P}(\sqrt{2}, 1)$Equation of tangent at $\mathrm{P}$$\sqrt{2} \cdot \mathrm{x}+\mathrm{y}=3$Let $\mathrm{Q}_{2}$ be $(0, \mathrm{k})$ and radius is $2 \sqrt{3}$$\therefore\left|\frac{\sqrt{2}(0)+k-3}{\sqrt{2+1}}\right|=2 \sqrt{3}$$\therefore \mathrm{k}=9,-3$$\mathrm{Q}_{2}(0,9)$ and $\mathrm{Q}_{3}(0,-3)$hence $\mathrm{Q}_{2} \mathrm{Q}_{3}=12$$\mathrm{R}_{2} \mathrm{R}_{3}$ is internal common tangent of circle $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$$\begin{aligned} \therefore \mathrm{R}_{2} \mathrm{R}_{3} &=\sqrt{\left(\mathrm{Q}_{2} \mathrm{Q}_{3}\right)^{2}-(2 \sqrt{3}+2 \sqrt{3})^{2}} \\ &=\sqrt{12^{2}-48}=\sqrt{96}=4 \sqrt{6} \end{aligned}$Perpendicular distance of origin $\mathrm{O}$ from $\mathrm{R}_{2} \mathrm{R}_{3}$ is equal to radius of circle $\mathrm{C}_{1}=\sqrt{3}$Hence area of $\Delta \mathrm{OR}_{2} \mathrm{R}_{3}=\frac{1}{2} \times\left(\mathrm{R}_{2} \mathrm{R}_{3}\right) \sqrt{3}=\frac{1}{2} \cdot 4 \sqrt{6} \cdot \sqrt{3}=6 \sqrt{2}$Perpendicular Distance of $\mathrm{P}$ from $\mathrm{Q}_{2} \mathrm{Q}_{3}=\sqrt{2}$$\therefore$ Area of $\Delta \mathrm{PQ}_{2} \mathrm{Q}_{3}=\frac{1}{2} \times 12 \times \sqrt{2}=6 \sqrt{2}$
Q. Let $P$ be the point on the parabola $y^{2}=4 x$ which is at the shortest distance from the center Sof the circle $x^{2}+y^{2}-4 x-16 y+64=0$. Let $Q$ be the point on the circle dividing the line segment SP internally. Then- (A) $\mathrm{SP}=2 \sqrt{5}$(B) $\mathrm{SQ}: \mathrm{QP}=(\sqrt{5}+1): 2$(C) the $\mathrm{x}$ -intercept of the normal to the parabola at $\mathrm{P}$ is 6(D) the slope of the tangent to the circle at $\mathrm{Q}$ is $\frac{1}{2}$[JEE (Advanced) 2016]
Ans. (A,C,D) point $P$ lies on normal to parabola passing through centre of circle$y+t x=2 t+t^{3}$$8+2 t=2 t+t^{3}$$t=2$$t=2$$P(4,4)$$S P=\sqrt{(4-2)^{2}+(4-8)^{2}}$$S P=2 \sqrt{5}$$\mathrm{SQ}=2$$\Rightarrow \mathrm{PQ}=2 \sqrt{5}-2$$\frac{\mathrm{SQ}}{\mathrm{QP}}=\frac{1}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{4}$To find $\mathrm{x}$ interceptput $\mathrm{y}=0 \mathrm{in}(\mathrm{i})$$\Rightarrow \mathrm{x}=2+\mathrm{t}^{2}$$\because=6$$\because \quad$ Slope of common normal $=-\mathrm{t}=-2$$\therefore \quad$ Slope of tangent $=\frac{1}{2}$
Q. If a chord, which is not a tangent, of the parabola $y^{2}=16 x$ has the equation $2 x+y=p,$ and midpoint $(h, k),$ then which of the following is (are) possible value(s) of $p, h$ and $k ?$ (A) p = 5, h = 4, k = –3(B) p = –1, h = 1, k = –3(C) p = –2, h = 2, k = –4(D) p = 2, h = 3, k = –4[JEE (Advanced) 2017]
Ans. (D) Equation of chord with mid point $(\mathrm{h}, \mathrm{k}):$$\mathrm{k} \cdot \mathrm{y}-16\left(\frac{\mathrm{x}+\mathrm{h}}{2}\right)=\mathrm{k}^{2}-16 \mathrm{h}$$\Rightarrow 8 \mathrm{x}-\mathrm{ky}+\mathrm{k}^{2}-8 \mathrm{h}=0$Comparing with $2 \mathrm{x}+\mathrm{y}-\mathrm{p}=0,$ we get$\mathrm{k}=-4 ; 2 \mathrm{h}-\mathrm{p}=4$only (D) satisfies above relation.
Share Blog
Share via...
Or copy link
Frequently Asked Questions
Find answers to common questions.
What are the most important formulas for parabola in JEE Advanced?
The five most frequently used formulas are: (1) Parametric form: (at², 2at) for y² = 4ax. (2) Equation of tangent at (at², 2at): ty = x + at². (3) Equation of normal: y = mx − 2am − am³. (4) Focal chord condition: t₁t₂ = −1. (5) Length of focal chord: a(t₁ − t₂)².
Is parabola important for JEE Advanced or just JEE Main?
Parabola is important for both, but JEE Advanced tests it at a deeper level. JEE Main focuses on standard formulas like equation of tangent and normal. JEE Advanced combines parabola with circles, loci, and geometric reasoning in multi-step problems requiring 4–6 logical jumps.
How many questions on parabola come in JEE Advanced?
Typically 1–3 questions per year appear on parabola in JEE Advanced, often as part of paragraph-based or multi-correct sets. Between 2009 and 2016, there were 16 questions totalling approximately 52 marks, making it one of the most heavily tested coordinate geometry topics.
Can parabola questions appear in JEE Advanced Paper 2 as well?
Yes. Parabola questions have appeared in both Paper 1 and Paper 2 of JEE Advanced. The 2013 paragraph on focal chords spanned both papers' difficulty profile. Always treat both papers as equally important for coordinate geometry preparation.
How long does it take to prepare parabola for JEE Advanced?
With focused preparation, 10–14 days is sufficient — 3 days on theory and standard results, 4 days on JEE Main-level problems, and 5–7 days on JEE Advanced previous year questions with full solution analysis. Students enrolled in eSaral's Kota-quality online batches typically complete this in structured weekly modules.
Comments
Nobita
July 2, 2021, 7:50 p.m.
2019 ,2020 ????
Dora
July 2, 2021, 7:49 p.m.
Good
Akhil
May 17, 2021, 2:27 p.m.
What about remaining latest year questions
YASH
May 2, 2021, 4:45 p.m.
GHATIYAAAAAAAAAAA
AAAAAAAA
AAA
A
AAAAAAAAAAAAAA
AAAAAAA
A
AAAAAAAAAAAAAAAAAAAAAAAAAA
Ratnakar
Feb. 15, 2021, 2:01 p.m.
Last year papers and with solution.
My wish
Oct. 9, 2020, 8:03 p.m.
It's nice but it will be better if you provide to put the answer
Shivu
Sept. 16, 2020, 12:34 a.m.
Nice question sir
Thank u
Gghjjj
July 4, 2020, 4:09 p.m.
Gyghhhh
saketh varma
July 4, 2020, 10:09 a.m.
sir rani mam is correct
S.Saketh varma
July 4, 2020, 10:08 a.m.
sir can i get the conept of parabola in clear manner
Rani
June 5, 2020, 7:25 p.m.
Not an updated one...
No much use in solving outdated models .... So if there is an updated report regarding previous years jee advanced questions ..... that makes it more useful...and also u can get benefitted..... bcauzz many will download ur app....
It's my suggestion which will benefit you and also the students who r aspiring for jee mains & jee advanced....
chord will intersect on directrix. $\therefore \quad \mathrm{M}\left(-\mathrm{a}, \mathrm{a}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)\right)$ lies on $\mathrm{y}=2 \mathrm{x}+\mathrm{a}$ $\mathrm{a}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)=-2 \mathrm{a}+\mathrm{a} \quad \Rightarrow \quad \mathrm{t}_{1}+\mathrm{t}_{2}=-1$ $\quad \& \quad \mathrm{t}_{1} \mathrm{t}_{2}=-1$ $\tan \theta=\left(\frac{\frac{2}{t_{1}}-\frac{2}{t_{2}}}{1+\frac{4}{t_{1} t_{2}}}\right)=\left(\frac{2\left(t_{2}-t_{1}\right)}{3}\right)$ $\because\left(t_{2}-t_{1}\right)=\left(t_{2}+t_{1}\right)^{2}-4 t_{1} t_{2}=5$ $t_{2}-t_{1}=\pm \sqrt{5}$ $\therefore \quad \tan \theta=\pm \frac{2 \sqrt{5}}{3}$ but $\theta$ is obtuse because $\mathrm{O}$ is the interior point of the circle for which $\mathrm{PQ}$ is diameter. $\therefore \quad \tan \theta=\frac{-2 \sqrt{5}}{3}$ 
[JEE(Advanced) 2013, 3, (–1)] 
$\mathrm{m}_{\mathrm{QR}}=\frac{2 \mathrm{a}\left(\mathrm{r}+\frac{1}{\mathrm{t}}\right)}{\mathrm{a}\left(\mathrm{r}^{2}-\frac{1}{\mathrm{t}^{2}}\right)}=\frac{2}{\left(\mathrm{r}-\frac{1}{\mathrm{t}}\right)}$ $\mathrm{m}_{\mathrm{PK}}=\frac{2 \mathrm{at}-0}{\mathrm{a}\left(\mathrm{t}^{2}-2\right)}=\frac{2 \mathrm{t}}{\mathrm{t}^{2}-2}$ Given $\mathrm{m}_{\mathrm{QR}}=\mathrm{m}_{\mathrm{PK}}$ $\Rightarrow \frac{2}{\mathrm{r}-\frac{1}{\mathrm{t}}}=\frac{2 \mathrm{t}}{\mathrm{t}^{2}-2} \Rightarrow \mathrm{r}=\frac{\mathrm{t}^{2}-2}{\mathrm{t}}+\frac{1}{\mathrm{t}}$ $\Rightarrow \mathrm{r}=\mathrm{t}-\frac{2}{\mathrm{t}}+\frac{1}{\mathrm{t}} \Rightarrow \mathrm{r}=\frac{\mathrm{t}^{2}-1}{\mathrm{t}}$ 
point $P$ lies on normal to parabola passing through centre of circle $y+t x=2 t+t^{3}$ $8+2 t=2 t+t^{3}$ $t=2$ $t=2$ $P(4,4)$ $S P=\sqrt{(4-2)^{2}+(4-8)^{2}}$ $S P=2 \sqrt{5}$ $\mathrm{SQ}=2$ $\Rightarrow \mathrm{PQ}=2 \sqrt{5}-2$ $\frac{\mathrm{SQ}}{\mathrm{QP}}=\frac{1}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{4}$ To find $\mathrm{x}$ intercept put $\mathrm{y}=0 \mathrm{in}(\mathrm{i})$ $\Rightarrow \mathrm{x}=2+\mathrm{t}^{2}$ $\because=6$ $\because \quad$ Slope of common normal $=-\mathrm{t}=-2$ $\therefore \quad$ Slope of tangent $=\frac{1}{2}$