Question

Let \(e_1\) and \(e_2\) be the eccentricities of the ellipse \(2x^2+9y^2=36\) and the hyperbola \(4x^2-9y^2=36\), respectively. Then the distance between the point of intersection of the lines \(5x-7y=3\) and \(3x+y=7\), and the point \( (9e_1^2,\;9e_2^2) \) is:

• A. \(11\)
• B. \(12\)
• C. \(13\)
• D. \(15\)

Hint:

Always convert conic equations to standard form first; eccentricities and related points follow directly.

Answer 1

The Correct Option is C

Concept:
Standard form of ellipse: \( \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 \), eccentricity \( e=\sqrt{1-\dfrac{b^2}{a^2}} \).
Standard form of hyperbola: \( \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 \), eccentricity \( e=\sqrt{1+\dfrac{b^2}{a^2}} \).
Distance between two points \( (x_1,y_1) \) and \( (x_2,y_2) \) is \( \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \).
Step 1: Find \(e_1\) for the ellipse \[ 2x^2+9y^2=36 \Rightarrow \frac{x^2}{18}+\frac{y^2}{4}=1 \] Thus, \(a^2=18,\ b^2=4\). \[ e_1=\sqrt{1-\frac{4}{18}}=\sqrt{\frac{14}{18}}=\frac{\sqrt7}{3} \]
Step 2: Find \(e_2\) for the hyperbola \[ 4x^2-9y^2=36 \Rightarrow \frac{x^2}{9}-\frac{y^2}{4}=1 \] Thus, \(a^2=9,\ b^2=4\). \[ e_2=\sqrt{1+\frac{4}{9}}=\sqrt{\frac{13}{9}}=\frac{\sqrt{13}}{3} \]
Step 3: Coordinates of the given point \[ (9e_1^2,9e_2^2)=\left(9\cdot\frac{7}{9},\,9\cdot\frac{13}{9}\right)=(7,13) \]
Step 4: Point of intersection of the lines From \(3x+y=7\), \(y=7-3x\). Substitute in \(5x-7y=3\): \[ 5x-7(7-3x)=3 \Rightarrow 26x=52 \Rightarrow x=2,\ y=1 \] So the point is \((2,1)\).
Step 5: Required distance \[ \sqrt{(7-2)^2+(13-1)^2}=\sqrt{25+144}=\sqrt{169}=13 \]