Question

In an ellipse $\frac{x^2}{10}+\frac{y^2}{6}=1$. If focal chords PSP' and QSQ' are at right angles to each other, then $\frac{1-e^2}{(SP)(SP')}+\frac{1-e^2}{(SQ)(SQ')}=\frac{N}{15}$. The value of [N] is (where [.] is greatest integer function)

• A. 4
• B. 5
• C. 3
• D. 2

Answer 1

Answer: (A)

4

Answer 2

The equation of the ellipse is $\frac{x^2}{10}+\frac{y^2}{6}=1$. Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we have $a^2 = 10$ and $b^2 = 6$.

The eccentricity $e$ is given by $b^2 = a^2(1-e^2)$, so $6 = 10(1-e^2)$, which gives $1-e^2 = \frac{6}{10} = \frac{3}{5}$. Thus, $e^2 = 1 - \frac{3}{5} = \frac{2}{5}$.

The semi-latus rectum $l$ is given by $l = \frac{b^2}{a} = \frac{6}{\sqrt{10}}$. So, $l^2 = \left(\frac{6}{\sqrt{10}}\right)^2 = \frac{36}{10} = \frac{18}{5}$.

For a focal chord, the reciprocal of the product of the segments from the focus is given by $\frac{1}{(SP)(SP')} = \frac{1-e^2 \cos^2 \theta}{l^2}$, where $\theta$ is the angle the chord makes with the major axis.

Let $PSP'$ make an angle $\theta$ with the major axis, and $QSQ'$ make an angle $\phi = \theta + \frac{\pi}{2}$ with the major axis. Then, $\frac{1}{(SP)(SP')} = \frac{1-e^2 \cos^2 \theta}{l^2}$ and $\frac{1}{(SQ)(SQ')} = \frac{1-e^2 \cos^2 (\theta + \frac{\pi}{2})}{l^2} = \frac{1-e^2 \sin^2 \theta}{l^2}$.

The given equation is $\frac{1-e^2}{(SP)(SP')}+\frac{1-e^2}{(SQ)(SQ')}=\frac{N}{15}$. Substituting the expressions: $(1-e^2) \left( \frac{1-e^2 \cos^2 \theta}{l^2} + \frac{1-e^2 \sin^2 \theta}{l^2} \right) = \frac{N}{15}$ $\frac{1-e^2}{l^2} \left( 1 - e^2 \cos^2 \theta + 1 - e^2 \sin^2 \theta \right) = \frac{N}{15}$ $\frac{1-e^2}{l^2} \left( 2 - e^2 (\cos^2 \theta + \sin^2 \theta) \right) = \frac{N}{15}$ $\frac{1-e^2}{l^2} (2 - e^2) = \frac{N}{15}$

Substitute the values $1-e^2 = \frac{3}{5}$, $e^2 = \frac{2}{5}$, and $l^2 = \frac{18}{5}$: $\frac{N}{15} = \frac{(\frac{3}{5})(2-\frac{2}{5})}{\frac{18}{5}} = \frac{(\frac{3}{5})(\frac{8}{5})}{\frac{18}{5}} = \frac{\frac{24}{25}}{\frac{18}{5}} = \frac{24}{25} \times \frac{5}{18} = \frac{24}{5 \times 18} = \frac{4}{15}$.

So, $\frac{N}{15} = \frac{4}{15}$, which implies $N = 4$. The value of $[N]$ is $[4] = 4$.