Question

If the distance between the foci of an ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] is \(6\) and the distance between its directrices is \(10\), then the equation of one of the tangents of the ellipse drawn parallel to the line \[ y=\sqrt2\,x+5 \] is

• A. \[ y=\sqrt2\,x+\sqrt{66} \]
• B. \[ y=\sqrt2\,x+12 \]
• C. \[ y=\sqrt2\,x+\sqrt{44} \]
• D. \[ y=\sqrt2\,x+6 \] \bigskip

Hint:

For the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] a tangent with slope \(m\) is \[ y=mx\pm\sqrt{a^2m^2+b^2}. \] Once \(a^2\) and \(b^2\) are known, the tangent can be written immediately.

Answer 1

The Correct Option is D

Concept: Use relationships between foci distance, directrix distance, and ellipse parameters. Then use tangent condition with given slope to find the tangent equation.

Step 1: Distance between foci = \(2c = 6 \Rightarrow c=3\). Distance between directrices = \(2a/e = 10 \Rightarrow a/e=5\).

Step 2: \(e=c/a \Rightarrow a/e = a^2/c = 5 \Rightarrow a^2=15\). \(c^2=a^2-b^2 \Rightarrow 9=15-b^2 \Rightarrow b^2=6\).

Step 3: Tangent parallel to \(y=\sqrt2 x+5\) has slope \(m=\sqrt2\). Tangent equation: \(y = mx \pm \sqrt{a^2m^2+b^2} = \sqrt2 x \pm \sqrt{15\cdot2 + 6} = \sqrt2 x \pm 6\).

Step 4: One such tangent is \(y = \sqrt2 x + 6\).

Step 5: Write the final answer. \(\boxed{y=\sqrt2 x+6}\)