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: Understanding the Concept
We need to find the midpoint of the chord that the line $y=x+1$ cuts on the ellipse $x^2+\frac{y^2}{4}=1$. This can be done by finding the two points of intersection and then using the midpoint formula.
Step 2: Key Formula or Approach
1. Rewrite the ellipse equation in a simpler form: $4x^2+y^2=4$. 2. Substitute the equation of the line ($y=x+1$) into the equation of the ellipse to get a quadratic equation in $x$. The roots of this quadratic, $x_1$ and $x_2$, will be the x-coordinates of the intersection points. 3. The x-coordinate of the midpoint is the average of the roots, $x_M = \frac{x_1+x_2}{2}$. We can find the sum of the roots using Vieta's formulas without solving for the roots individually. 4. Once $x_M$ is found, the y-coordinate $y_M$ can be found by substituting $x_M$ into the line equation, since the midpoint lies on the chord.
Step 3: Detailed Explanation
1. Set up the equations: Ellipse: $x^2+\frac{y^2}{4}=1 \implies 4x^2+y^2=4$. Line (Chord): $y=x+1$. 2. Find the intersection points' x-coordinates: Substitute the line equation into the ellipse equation: \[ 4x^2 + (x+1)^2 = 4 \] \[ 4x^2 + (x^2+2x+1) = 4 \] \[ 5x^2 + 2x + 1 - 4 = 0 \] \[ 5x^2 + 2x - 3 = 0 \] This is a quadratic equation whose roots, $x_1$ and $x_2$, are the x-coordinates of the endpoints of the chord. 3. Find the x-coordinate of the midpoint: Let the midpoint be $(x_M, y_M)$. \[ x_M = \frac{x_1+x_2}{2} \] From the quadratic equation $5x^2+2x-3=0$, the sum of the roots is $x_1+x_2 = -\frac{b}{a} = -\frac{2}{5}$. \[ x_M = \frac{-2/5}{2} = -\frac{1}{5} \] 4. Find the y-coordinate of the midpoint: The midpoint lies on the line $y=x+1$. \[ y_M = x_M + 1 = -\frac{1}{5} + 1 = \frac{-1+5}{5} = \frac{4}{5} \] Step 4: Final Answer
The midpoint of the chord is $\left(-\frac{1}{5}, \frac{4}{5}\right)$.