Let S and S^{prime} be the foci of the ellips | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$, we have $a^2=25$ and $b^2=9$, so $a=5$ and $b=3$. Eccentricity $e = \sqrt{1-\frac{b^2}{a^2}} =

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(D) Given the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$, we have $a^2=25$ and $b^2=9$, so $a=5$ and $b=3$. Eccentricity $e = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{9}{25}} = \frac{4}{5}$. The foci are $S(ae, 0) = (4, 0)$ and $S^{\prime}(-ae, 0) = (-4, 0)$. Since $P(\alpha, \beta)$ lies on the ellipse, $SP+S^{\prime}P=2a=10$. Given $(SP)^2+(S^{\prime}P)^2-SP \cdot S^{\prime}P=37$. Using $(SP+S^{\prime}P)^2 = (SP)^2+(S^{\prime}P)^2+2SP \cdot S^{\prime}P$, we substitute $(SP)^2+(S^{\prime}P)^2 = (SP+S^{\prime}P)^2-2SP \cdot S^{\prime}P$. So, $(SP+S^{\prime}P)^2-3SP \cdot S^{\prime}P=37$. $10^2-3SP \cdot S^{\prime}P=37$ $\Rightarrow 100-3SP \cdot S^{\prime}P=37$ $\Rightarrow 3SP \cdot S^{\prime}P=63$ $\Rightarrow SP \cdot S^{\prime}P=21$. For any point $P(\alpha, \beta)$ on the ellipse, $SP = a-e\alpha$ and $S^{\prime}P = a+e\alpha$. $SP \cdot S^{\prime}P = a^2-e^2\alpha^2 = 25-\frac{16}{25}\alpha^2 = 21$. $\frac{16}{25}\alpha^2 = 4 \Rightarrow \alpha^2 = \frac{100}{16} = \frac{25}{4}$. From $\frac{\alpha^2}{25}+\frac{\beta^2}{9}=1$, we have $\frac{25/4}{25}+\frac{\beta^2}{9}=1$ $\Rightarrow \frac{1}{4}+\frac{\beta^2}{9}=1$ $\Rightarrow \frac{\beta^2}{9}=\frac{3}{4}$ $\Rightarrow \beta^2=\frac{27}{4}$. Thus, $\alpha^2+\beta^2 = \frac{25}{4}+\frac{27}{4} = \frac{52}{4} = 13$.

Let $S$ and $S^{\prime}$ be the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ and $P(\alpha, \beta)$ be a point on the ellipse in the first quadrant. If $(SP)^2+(S^{\prime}P)^2-SP \cdot S^{\prime}P=37$, then $\alpha^2+\beta^2$ is equal to:

Similar Questions

If a focal chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ meets its minor axis at the point $(0,3)$,then the perpendicular distance from the centre of the ellipse to this focal chord is

The minimum area of a triangle formed by any tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{81} = 1$ and the coordinate axes is

Find the equation for the ellipse that satisfies the given conditions: Major axis on the $x-$ axis and passes through the points $(4, 3)$ and $(6, 2)$.

The number of tangents to the circle $x^2 + y^2 = 3$ that are normal to the ellipse $4x^2 + 9y^2 = 36$ is

What is the equation of the chord of the ellipse $\frac{x^2}{36} + \frac{y^2}{9} = 1$ that is bisected at the point $(2, 1)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module