An ellipse with axes parallel to coordinate axes cuts the parabola $y^2 = 4x$ at $(1, –2)$ and touches it at $(4, 4)$ then the coordinate of other point of intersection is
(A) $(4, –4)$
(B) $(3, –2)$
(C) $(3, 2)$
(D) $(9, –6)$
The diagram is illustrated above where in the ellipse touches at $(4,4)$ and cuts the ellipse at $(1,-2)$.
For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $a>b$ we can apply the property $PA+PB=2a$ where A nd B are focus and P is any point on the ellipse but not able to solve this problem
Let the centre of the ellipse be $(h,k)$ and as it's a classic case of a shifted ellipse, it's equation can be written as $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$, Solving the ellipse and parabola by substituting x as $\frac{y^2}{4}$ in ellipse equation you get a biquadratic equation in y, whose sum of roots say $y_1$+$y_2$+$y_3$+$y_4$ = 0, where $y_1$ = $y_2$ = 4 and $y_3$ = -2 and assuming the fourth point of intersection as $(\alpha,\beta)$; $y_4$ = $\beta$ you get, $\beta$ = -6 and putting $\beta$ in the parabola equation you get $\alpha$ = 9, hence $(9,-6)$