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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) $(9, –6)$

enter image description here

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

Samar Imam Zaidi
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    i suppose from figure you could say option D to be right, also only A and D satisfy the parabola – NadiKeUssPar May 07 '23 at 07:32
  • ... and my intuition is that given the ellipse has an axis parallel to the x axis, that the parabola is symmetric across the x axis, and that the ellipse and parabola are tangent at (4, 4), if the two intersected at (4, -4), the reflection of (4, 4) through the x axis, then they would also be tangent there. – John Bollinger May 07 '23 at 16:36

1 Answers1

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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 $(9,-6)$

NadiKeUssPar
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