Question

The tangents drawn at the points \(P_1\) and \(P_2\) lying on the ellipse \(\frac{x^2}{4}+y^2=1\) are parallel to the chord joining the points \((0,1)\) and \((2,0)\), then the distance between \(P_1\) and \(P_2\) is:

• A. \(2\sqrt2\)
• B. \(\sqrt5\)
• C. \(2\sqrt3\)
• D. \(\sqrt{10}\)

Hint:

For ellipse tangent problems, compare tangent slope with the required line slope and then substitute back into the ellipse equation.

Answer 1

The Correct Option is D

Concept: For ellipse: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] the tangent at \((x_1,y_1)\) is: \[ \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1 \] Its slope is: \[ m=-\frac{b^2x_1}{a^2y_1} \]

Step 1: Find slope of given chord.
Points are: \[ (0,1),\quad (2,0) \] Slope: \[ m=\frac{0-1}{2-0}=-\frac12 \]

Step 2: Equation of ellipse.
Given ellipse: \[ \frac{x^2}{4}+y^2=1 \] Hence, \[ a^2=4,\quad b^2=1 \] Slope of tangent at \((x_1,y_1)\): \[ m=-\frac{x_1}{4y_1} \] Set equal to \(-\frac12\): \[ -\frac{x_1}{4y_1}=-\frac12 \] \[ x_1=2y_1 \]

Step 3: Substitute into ellipse equation.
\[ \frac{(2y_1)^2}{4}+y_1^2=1 \] \[ y_1^2+y_1^2=1 \] \[ 2y_1^2=1 \] \[ y_1=\pm\frac1{\sqrt2} \] Thus, \[ x_1=\pm\sqrt2 \] The two points are: \[ (\sqrt2,\tfrac1{\sqrt2}) \] and \[ (-\sqrt2,-\tfrac1{\sqrt2}) \]

Step 4: Find distance between points.
\[ d=\sqrt{(2\sqrt2)^2+\left(\frac2{\sqrt2}\right)^2} \] \[ =\sqrt{8+2} \] \[ =\sqrt{10} \] Hence the answer is: \[ \boxed{\sqrt{10}} \]