If the coordinates of two points $A$ and $B$ are $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$ respectively and $P$ is any point on the conic $9x^{2} + 16y^{2} = 144$,then $PA + PB$ is equal to

  • A
    $8$
  • B
    $6$
  • C
    $16$
  • D
    $9$

Explore More

Similar Questions

Let $A, A^{\prime}$ be the end points of the major axis,$S, S^{\prime}$ be the foci,and $B, B^{\prime}$ be the end points of the minor axis of an ellipse $E$. If $\angle BAB^{\prime}=60^{\circ}$,then find $\angle SBS^{\prime}$.

The area (in sq. units) of the triangle formed by the tangent and normal at a point $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$ to the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the $X$-axis is

The eccentricity of the ellipse $\left( \frac{x - 3}{y} \right)^2 + \left( 1 - \frac{4}{y} \right)^2 = \frac{1}{9}$ is

Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$-axis and the $y$-axis,respectively. Let $S$ be the circle $x^2+(y-1)^2=2$. The straight line $x+y=3$ touches the curves $S, E_1$ and $E_2$ at $P, Q$ and $R$,respectively. Suppose that $PQ=PR=\frac{2 \sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$,respectively,then the correct expression$(s)$ is(are):
$(A) e_1^2+e_2^2=\frac{43}{40}$
$(B) e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$
$(C) |e_1^2-e_2^2|=\frac{5}{8}$
$(D) e_1 e_2=\frac{\sqrt{3}}{4}$

If $C$ is the centre of the ellipse $9x^2 + 16y^2 = 144$ and $S$ is one focus,then the ratio of $CS$ to the major axis is:

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