Parabola – JEE Main Previous Year Question with Solutions
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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]

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Sol. (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]

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Sol. (C,D)$\mathrm{t}_{1}+\mathrm{t}_{2}=\mathrm{r}$$\frac{2}{\mathrm{r}}=\frac{2}{\mathrm{t}_{1}+\mathrm{t}_{2}}similarly -\frac{2}{\mathrm{r}} isalso possible Q. Consider the parabola \mathrm{y}^{2}=8 \mathrm{x}. Let \Delta_{1} be the area of the triangle formed by the end points of its latus rectum and the point \mathrm{P}\left(\frac{1}{2}, 2\right) on the parabola, and \Delta_{2} be the area of the triangle formed by drawing tangents at \mathrm{P} and at the end points of the latus rectum. Then \frac{\Delta_{1}}{\Delta_{2}} is [JEE 2011,4] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 2  Q. Let (x,y) be any point on the parabola \mathrm{y}^{2} = 4x. Let P be the point that divides the line segment from (0,0) to (x,y) in the ratio 1 : 3. Then the locus of P is-(A) x^{2}=y(B) \mathrm{y}^{2}=2 \mathrm{x}(C) \mathrm{y}^{2}=\mathrm{x}(D) x^{2}=2 y [JEE 2011,3] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (C)Let P be (h, k)on using section formula \mathrm{P}\left(\frac{x}{4}, \frac{y}{4}\right)$$\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}$$\therefore \quad 16 \mathrm{k}^{2}=16 \mathrm{h} \quad \Rightarrow \mathrm{k}^{2}=\mathrm{h}Locus of point \mathrm{P} is \mathrm{y}^{2}=\mathrm{x} Q. Let L be a normal to the parabola \mathrm{y}^{2}=4 \mathrm{x} . If \mathrm{L} passes through the point (9,6), then \mathrm{L} is given by -(A) y – x + 3 =0(B) y + 3x – 33 = 0(C) y + x – 15 = 0(D) y – 2x + 12 = 0 [JEE 2011,3] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,B,D)Equation of normal is \mathrm{y}=\mathrm{mx}-2 \mathrm{m}-\mathrm{m}^{3}It passes through the point (9,6) then6=9 \mathrm{m}-2 \mathrm{m}-\mathrm{m}^{3}$$\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]

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Sol. 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)]

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Sol. (D)Single tangent at the extrimities of a focal 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. $\therefore \quad \tan \theta=\frac{-2 \sqrt{5}}{3}$

Q. Length of chord PQ is(A) 7a             (B) 5a            (C) 2a             (D) 3a [JEE(Advanced) 2013, 3, (–1)]

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Sol. (B)\begin{aligned} \text { Length of focal chord } \\ \mathrm{PQ} &=\mathrm{a}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)^{2} \\ &=\mathrm{a}\left[\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}-4 \mathrm{t}_{1} \mathrm{t}_{2}\right] \\ &=\mathrm{a}[1+4]=5 \mathrm{a} \end{aligned}

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$$\leq 6 at the point F\left(x_{0}, y_{0}\right). The tangent to the parabola at F\left(x_{0}, y_{0}\right) intersects the y -axis at G\left(0, y_{1}\right) .The slope m of the line L is chosen such that the area of the triangle EFG has a local maximum.Match List-I with List-II and select the correct answer using the code given below the lists.  [JEE(Advanced) 2013, 3, (–1)] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A )  Q. The common tangents to the circle \mathrm{x}^{2}+\mathrm{y}^{2}=2 and the parabola \mathrm{y}^{2}=8 \mathrm{x} touch the circle at the point \mathrm{P}, \mathrm{Q} and the parabola at the points \mathrm{R}, \mathrm{S}. Then the area of the quadrilateral PQRSis –(A) 3 (B) 6 (C) 9 (D) 15 [JEE(Advanced)-2014, 3(–1)] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (D)  Paragraph For Questions 11 and 12Let a,r,s, t be nonzero real numbers. Let P\left(a t^{2}, 2 a t\right), Q, R\left(a r^{2}, 2 a r\right) and S\left(a s^{2}, 2 a s\right) be distinct points on the parabola y^{2}=4 a x . Suppose that P Q is the focal chord and lines Q R and P K are parallel, where K is the point (2 a, 0) .The value of r is- Q. The value of r is-(A) -\frac{1}{\mathfrak{t}}(B) \frac{t^{2}+1}{t}(C) \frac{1}{\mathrm{t}}(D) \frac{\mathfrak{t}^{2}-1}{\mathfrak{t}} [JEE(Advanced)-2014, 3(–1)] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (D)\because \mathrm{PQ} is a focal chord\therefore co-ordinates of point \mathrm{Q} are =\left(\frac{\mathrm{a}}{\mathrm{t}^{2}},-\frac{2 \mathrm{a}}{\mathrm{t}}\right) \mathrm{m}_{\mathrm{QR}}=\frac{2 \mathrm{a}\left(\mathrm{r}+\frac{1}{\mathrm{t}}\right)}{\mathrm{a}\left(\mathrm{r}^{2}-\frac{1}{\mathrm{t}^{2}}\right)}=\frac{2}{\left(\mathrm{r}-\frac{1}{\mathrm{t}}\right)}$$\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)]

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Sol. (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]

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Sol. 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]

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Sol. 4Let 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]

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Sol. (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]

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Sol. (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}$$\sqrt{2} \cdot \mathrm{x}+\mathrm{y}=3Let \mathrm{Q}_{2} be (0, \mathrm{k}) and radius is 2 \sqrt{3}$$\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}$$\therefore Area of \Delta \mathrm{PQ}_{2} \mathrm{Q}_{3}=\frac{1}{2} \times 12 \times \sqrt{2}=6 \sqrt{2} Q. Let P be the point on the parabola y^{2}=4 x which is at the shortest distance from the center Sof the circle x^{2}+y^{2}-4 x-16 y+64=0. Let Q be the point on the circle dividing the line segment SP internally. Then-(A) \mathrm{SP}=2 \sqrt{5}(B) \mathrm{SQ}: \mathrm{QP}=(\sqrt{5}+1): 2(C) the \mathrm{x} -intercept of the normal to the parabola at \mathrm{P} is 6(D) the slope of the tangent to the circle at \mathrm{Q} is \frac{1}{2} [JEE (Advanced) 2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,C,D) point P lies on normal to parabola passing through centre of circley+t x=2 t+t^{3}$$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]

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Sol. (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.