Q.12. Let be a directrix of an ellipse E, whose centre is at the origin and eccentricity is . Let ,

Q: What is the correct answer for JEE Main Mathematics Q12?

The correct answer is A. Sol. For ellipse E: Center at origin $(0,0)$. Eccentricity $e = \frac{1}{3}$. Directrix is $x=9$. For an ellipse with center at origin, the directrix is $x = \frac{a}{e}$. So $\frac{a}{1/3} = 9$, which means $3a=9$, so $a=3$. The focus $P(\alpha, 0)$ with $\alpha > 0$ is $(ae, 0)$. So $\alpha = ae = 3 \cdot \frac{1}{3} = 1$. Thus, the focus is $P(1,0)$. The equation of the ellipse is $\frac{x^2}{a

Q: Q.12. Let be a directrix of an ellipse E, whose centre is at the origin and eccentricity is . Let , , be a focus of E and AB be a chord passing through P. Then the locus of the mid point of AB is :

The correct answer is A. Sol. For ellipse E: Center at origin $(0,0)$. Eccentricity $e = \frac{1}{3}$. Directrix is $x=9$. For an ellipse with center at origin, the directrix is $x = \frac{a}{e}$. So $\frac{a}{1/3} = 9$, which means $3a=9$, so $a=3$. The focus $P(\alpha, 0)$ with $\alpha > 0$ is $(ae, 0)$. So $\alpha = ae = 3 \cdot \frac{1}{3} = 1$. Thus, the focus is $P(1,0)$. The equation of the ellipse is $\frac{x^2}{a

Q: What are the options for this JEE Main Mathematics question?

A. $9y^2 = 8x(1-x)$ · B. $3y^2 = 4x(1-x)$ · C. $9y^2 = 8x(x-1)$ · D. $3y^2 = 4x(x-1)$

Question

Q.12. Let  be a directrix of an ellipse E, whose centre is at the origin and eccentricity is . Let , , be a focus of E and AB be a chord passing through P. Then the locus of the mid point of AB is :

FirstInTest · Accepted Answer

Sol. For ellipse E: Center at origin $(0,0)$. Eccentricity $e = \frac{1}{3}$. Directrix is $x=9$. For an ellipse with center at origin, the directrix is $x = \frac{a}{e}$. So $\frac{a}{1/3} = 9$, which means $3a=9$, so $a=3$. The focus $P(\alpha, 0)$ with $\alpha > 0$ is $(ae, 0)$. So $\alpha = ae = 3 \cdot \frac{1}{3} = 1$. Thus, the focus is $P(1,0)$. The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. We have $b^2 = a^2(1-e^2) = 3^2(1 - (\frac{1}{3})^2) = 9(1 - \frac{1}{9}) = 9(\frac{8}{9}) = 8$. So the ellipse equation is $\frac{x^2}{9} + \frac{y^2}{8} = 1$. Let the chord AB pass through the focus $P(1,0)$. Let the midpoint of the chord AB be $(h,k)$. The equation of the chord whose midpoint is $(h,k)$ is $T=S_1$. $T = \frac{xh}{9} + \frac{yk}{8} - 1$. $S_1 = \frac{h^2}{9} + \frac{k^2}{8} - 1$. So the equation of the chord is $\frac{xh}{9} + \frac{yk}{8} - 1 = \frac{h^2}{9} + \frac{k^2}{8} - 1$. $\frac{xh}{9} + \frac{yk}{8} = \frac{h^2}{9} + \frac{k^2}{8}$. Since the chord passes through the focus $P(1,0)$, we substitute $(x,y) = (1,0)$ into the chord equation. $\frac{1 \cdot h}{9} + \frac{0 \cdot k}{8} = \frac{h^2}{9} + \frac{k^2}{8}$. $\frac{h}{9} = \frac{h^2}{9} + \frac{k^2}{8}$. Multiply by 72: $8h = 8h^2 + 9k^2$. $9k^2 = 8h - 8h^2 = 8h(1-h)$. Replacing $(h,k)$ with $(x,y)$, the locus of the midpoint is $9y^2 = 8x(1-x)$.

Answer

$9y^2 = 8x(1-x)$

Answer

$3y^2 = 4x(1-x)$

Answer

$9y^2 = 8x(x-1)$

Answer

$3y^2 = 4x(x-1)$

Related Questionsin Mathematics

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions