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(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]
(A) $\frac{-1}{\mathrm{r}}$ (B) $\frac{1}{\mathrm{r}}$ (C) $\frac{2}{\mathrm{r}}$ (D) $\frac{-2}{\mathrm{r}}$ [JEE 2010,3]
$\mathrm{t}_{1}+\mathrm{t}_{2}=\mathrm{r}$ $\frac{2}{\mathrm{r}}=\frac{2}{\mathrm{t}_{1}+\mathrm{t}_{2}}$ similarly $-\frac{2}{\mathrm{r}}$ is also possible
[JEE 2011,4]
(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]
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 $\mathrm{y}^{2}=\mathrm{x}$
(A) y – x + 3 =0 (B) y + 3x – 33 = 0 (C) y + x – 15 = 0 (D) y – 2x + 12 = 0 [JEE 2011,3]
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 $\mathrm{y}-\mathrm{x}+3=0, \mathrm{y}+3 \mathrm{x}-33=0$ & $\quad \mathrm{y}-2 \mathrm{x}+12=0$
[JEE 2012, 4M]
Paragraph for Question 7 and 8
Let 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 $\mathrm{y}=2 \mathrm{x}+\mathrm{a}, \mathrm{a}>0 .$
(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)]
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. $\therefore \quad \tan \theta=\frac{-2 \sqrt{5}}{3}$
(A) 7a (B) 5a (C) 2a (D) 3a [JEE(Advanced) 2013, 3, (–1)]
$\begin{aligned} \text { Length of focal chord } \\ \mathrm{PQ} &=\mathrm{a}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)^{2} \\ &=\mathrm{a}\left[\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}-4 \mathrm{t}_{1} \mathrm{t}_{2}\right] \\ &=\mathrm{a}[1+4]=5 \mathrm{a} \end{aligned}$
$\leq 6$ 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)]
is – (A) 3 (B) 6 (C) 9 (D) 15 [JEE(Advanced)-2014, 3(–1)]
Paragraph For Questions 11 and 12
Let 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-
(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)]
$\because \mathrm{PQ}$ 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 $\mathrm{m}_{\mathrm{QR}}=\mathrm{m}_{\mathrm{PK}}$ $\Rightarrow \frac{2}{\mathrm{r}-\frac{1}{\mathrm{t}}}=\frac{2 \mathrm{t}}{\mathrm{t}^{2}-2} \Rightarrow \mathrm{r}=\frac{\mathrm{t}^{2}-2}{\mathrm{t}}+\frac{1}{\mathrm{t}}$ $\Rightarrow \mathrm{r}=\mathrm{t}-\frac{2}{\mathrm{t}}+\frac{1}{\mathrm{t}} \Rightarrow \mathrm{r}=\frac{\mathrm{t}^{2}-1}{\mathrm{t}}$
(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)]
[JEE 2015, 4M, –0M]
[JEE 2015, 4M, –0M]
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 $\mathrm{A} \& \mathrm{B}=4$
(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]
(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]
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 $\Delta \mathrm{PQ}_{2} \mathrm{Q}_{3}=\frac{1}{2} \times 12 \times \sqrt{2}=6 \sqrt{2}$
(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]
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 $=\frac{1}{2}$
(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]
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.
Really helpful.. Thanks for the PYQs..