Question

The midpoint of the chord of the ellipse $x^2+\frac{y^2}{4}=1$ formed on the line $y=x+1$ is

• A. $\left(\frac{4}{5}, \frac{9}{5}\right)$
• B. $\left(-\frac{1}{5}, \frac{4}{5}\right)$
• C. $\left(-\frac{1}{5}, \frac{6}{5}\right)$
• D. $\left(\frac{1}{5}, \frac{6}{5}\right)$

Hint:

An alternative method uses the property that the equation of a chord of a conic with a given midpoint $(x_1, y_1)$ is given by $T=S_1$. For the ellipse $4x^2+y^2-4=0$, this would be $4xx_1+yy_1-4 = 4x_1^2+y_1^2-4$. The slope of this chord is $-4x_1/y_1$. Since the chord is $y=x+1$, its slope is 1. So, $-4x_1/y_1=1 \implies y_1=-4x_1$. Since $(x_1,y_1)$ is on the line, $y_1=x_1+1$. Solving these two gives $-4x_1=x_1+1 \implies -5x_1=1 \implies x_1=-1/5$, and $y_1=4/5$.

Answer 1

The Correct Option is B

Step 1: Equation of Chord with Midpoint: Let the midpoint be \( (h,k) \). Since it lies on the line \( y=x+1 \), we have \( k = h+1 \). The equation of the chord with midpoint \( (h,k) \) for ellipse \( S=0 \) is \( T=S_1 \). \( S = x^2 + \frac{y^2}{4} - 1 \). \( T = xh + \frac{yk}{4} - 1 \). \( S_1 = h^2 + \frac{k^2}{4} - 1 \). Equation: \( xh + \frac{yk}{4} = h^2 + \frac{k^2}{4} \).
Step 2: Compare with Given Line: The chord is given as \( y = x+1 \implies x - y = -1 \). Comparing coefficients of \( xh + \frac{k}{4}y = \text{const} \) with \( x - y = -1 \): \[ \frac{h}{1} = \frac{k/4}{-1} \] \[ h = -\frac{k}{4} \implies k = -4h \]
Step 3: Solve for h and k: Substitute \( k = -4h \) into \( k = h+1 \): \[ -4h = h+1 \] \[ -5h = 1 \implies h = -\frac{1}{5} \] Then \( k = -4\left(-\frac{1}{5}\right) = \frac{4}{5} \). Midpoint is \( \left(-\frac{1}{5}, \frac{4}{5}\right) \).