Question

A hyperbola intersects an ellipse $x^2 + 9y^2 = 9$ orthogonally. The eccentricity of the hyperbola is reciprocal of that of ellipse. If the axes of the hyperbola are along coordinate axes, then

• A. vertices of hyperbola are $(\pm \frac{8}{3}, 0)$
• B. y coordinate of point of intersection of ellipse and hyperbola is either $\frac{1}{3}$ or $-\frac{1}{3}$
• C. latus rectum of hyperbola is $\frac{2}{3}$
• D. latus rectum of hyperbola is $\frac{4}{3}$

Answer 1

Answer: (A), (B), (C)

Options (A), (B), and (C)

Answer 2

1. Ellipse:
   The given ellipse is

   $$x^2+9y^2=9 \quad\Longrightarrow\quad \frac{x^2}{9}+\frac{y^2}{1}=1.$$

   So $a=3$, $b=1$ and its eccentricity is

   $$e_{\text{ellipse}}=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{2}}{3}.$$

2. Hyperbola:
   Its eccentricity is given as the reciprocal of that of the ellipse

   $$e_{\text{hyp}}=\frac{3}{2\sqrt{2}}.$$

   For a hyperbola with axes along the coordinate axes and equation

   $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$$

   the eccentricity is

   $$e_{\text{hyp}}=\sqrt{1+\frac{b^2}{a^2}}.$$

   Equate:

   $$\sqrt{1+\frac{b^2}{a^2}}=\frac{3}{2\sqrt{2}}\quad\Longrightarrow\quad 1+\frac{b^2}{a^2}=\frac{9}{8}\quad\Longrightarrow\quad \frac{b^2}{a^2}=\frac{1}{8}.$$

   Hence,

   $$b^2=\frac{a^2}{8}.$$

3. Orthogonality Condition: At a common point $(x,y)$, the product of slopes of tangents to the ellipse and hyperbola must be $-1$.

   • For the ellipse, $x^2+9y^2=9$:
     Differentiating:

     $$2x+18yy'=0\quad\Longrightarrow\quad y'=-\frac{x}{9y}.$$

   • For the hyperbola, first write its equation as

     $$\frac{x^2}{a^2}-\frac{y^2}{(a^2/8)}=1 \quad\Longrightarrow\quad x^2-8y^2=a^2.$$

     Differentiating:

     $$2x-16yy'=0\quad\Longrightarrow\quad y'=\frac{x}{8y}.$$

   Orthogonality requires:

   $$\left(-\frac{x}{9y}\right)\left(\frac{x}{8y}\right)=-1\quad\Longrightarrow\quad \frac{x^2}{72y^2}=1 \quad\Longrightarrow\quad x^2=72y^2.$$

   Substitute $x^2=72y^2$ into the ellipse equation:

   $$72y^2+9y^2=9\quad\Longrightarrow\quad 81y^2=9\quad\Longrightarrow\quad y^2=\frac{1}{9}\quad\Longrightarrow\quad y=\pm\frac{1}{3}.$$

4. Finding $a$ (and vertices):
   Choose a point of intersection and substitute back into the hyperbola. Using $x^2=72y^2$ with $y^2=\frac{1}{9}$:

   $$x^2=72\left(\frac{1}{9}\right)=8.$$

   Now using the hyperbola equation $x^2-8y^2=a^2$:

   $$a^2=8-8\left(\frac{1}{9}\right)=8\left(1-\frac{1}{9}\right)=8\cdot\frac{8}{9}=\frac{64}{9}.$$

   So,

   $$a=\frac{8}{3}.$$

   Thus, the vertices are $\left(\pm\frac{8}{3},0\right)$.

5. Latus Rectum of the Hyperbola:
   For the hyperbola, the latus rectum length is given by:

   $$L=\frac{2b^2}{a}.$$

   We already have $b^2=\frac{a^2}{8}=\frac{64/9}{8}=\frac{8}{9}$. So,

   $$L=\frac{2\cdot\frac{8}{9}}{\frac{8}{3}}=\frac{16/9}{8/3}=\frac{16}{9}\cdot\frac{3}{8}=\frac{48}{72}=\frac{2}{3}.$$