Let the tangent drawn to the parabola $y^2 = 24x$ at the point $(\alpha, \beta)$ be perpendicular to the line $2x + 2y = 5$. Then the normal to the hyperbola $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$ at the point $(\alpha + 4, \beta + 4)$ does $NOT$ pass through which of the following points?

  • A
    $(25, 10)$
  • B
    $(20, 12)$
  • C
    $(30, 8)$
  • D
    $(15, 13)$

Explore More

Similar Questions

The Cartesian form of the curve given by $x = \frac{a}{2} (t + \frac{1}{t})$ and $y = \frac{a}{2} (t - \frac{1}{t})$,where $t$ is a parameter,is:

Let $S$ denote the locus of the point of intersection of the pair of lines $4x - 3y = 12\alpha$ and $4\alpha x + 3\alpha y = 12$, where $\alpha$ varies over the set of non-zero real numbers. Let $T$ be the tangent to $S$ passing through the points $(p, 0)$ and $(0, q)$, with $q > 0$, and parallel to the line $4x - \frac{3}{\sqrt{2}}y = 0$. Then the value of $pq$ is (in $\sqrt{2}$)

The distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt{2}$. Its equation is

If the latus rectum through one of the foci of a hyperbola $\frac{x^2}{9}-\frac{y^2}{b^2}=1$ subtends a right angle at the farther vertex of the hyperbola,then $b^2=$

If $e_1$ and $e_2$ are respectively the eccentricities of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate hyperbola,then the line $\frac{x}{2 e_1}+\frac{y}{2 e_2}=1$ touches the circle having centre at the origin. Find its radius.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo