Ellipse Question - None

Question

Question nos. 649 to 651
Consider, $E : \dfrac{(x-1)^2}{16} + \dfrac{(y-2)^2}{9} = 1$ and $H : (x-1)^2 - (y-2)^2 = \dfrac{7}{2}$.

Column-1 contains equation of tangent to either $E$ or $H$.
Column-2 contains image of foci (whose abscissa is greater than 1) of the conic in its tangent.
Column-3 contains area (in sq. units) of the triangle formed by joining foci of the conic (according to column-2), its image in the tangent and centre of the conic.

Column-1	Column-2	Column-3
(I) $y = x + 6$	(i) $(1, \sqrt{7}+2)$	(P) $\dfrac{7}{2}$
(II) $y = x + 1$	(ii) $(-4, \sqrt{7}+7)$	(Q) $\dfrac{5\sqrt{7}+7}{2}$
(III) $x + y = 3$	(iii) $(6, \sqrt{7}-3)$	(R) $\dfrac{7}{4}$
(IV) $x - y - 4 = 0$	(iv) $(1, 2-\sqrt{7})$	(S) $\dfrac{5\sqrt{7}-7}{2}$

Which of the following options is the only correct combination?

Accepted Answer

Step 1: The equation $E : \dfrac{(x-1)^2}{16} + \dfrac{(y-2)^2}{9} = 1$ represents an ellipse with center $(1, 2)$, semi-major axis $4$, and semi-minor axis $3$. The foci of the ellipse are given by $(1 \pm \sqrt{16-9}, 2) = (1 \pm \sqrt{7}, 2)$. The equation $H : (x-1)^2 - (y-2)^2 = \dfrac{7}{2}$ represents a hyperbola with center $(1, 2)$, and its asymptotes are $y = \pm \dfrac{\sqrt{7}}{2}(x-1) + 2$.

Step 2: To find the equation of the tangent to the ellipse or hyperbola, we need to find the slope of the tangent line. The slope of the tangent line to the ellipse at point $(x_1, y_1)$ is given by $-\dfrac{9(x_1-1)}{16(y_1-2)}$. The slope of the tangent line to the hyperbola at point $(x_1, y_1)$ is given by $\dfrac{(x_1-1)}{(y_1-2)}$. We can use these slopes to find the equation of the tangent line.

Step 3: The image of the foci in the tangent line can be found by using the reflection formula. If the tangent line is $y = mx + c$, then the image of the focus $(x_1, y_1)$ is given by $(x_2, y_2)$, where $x_2 = \dfrac{(1-m^2)x_1 + 2m(y_1-c)}{1+m^2}$ and $y_2 = \dfrac{2mx_1 - (1-m^2)y_1 + 2mc}{1+m^2}$. We can use this formula to find the image of the foci in the tangent line.

Step 4: The area of the triangle formed by joining the foci, its image in the tangent, and the center of the conic can be found by using the formula for the area of a triangle. The area of the triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by $\dfrac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$. We can use this formula to find the area of the triangle.

Step 5: By comparing the options, we can see that option (d) is the only correct combination. The equation of the tangent line is $x - y - 4 = 0$, the image of the focus is $(6, \sqrt{7}-3)$, and the area of the triangle is $\dfrac{7}{4}$.

Answer: (d)

Key Concept: The key concept used in this solution is the reflection formula, which is used to find the image of the foci in the tangent line. The reflection formula is given by $(x_2, y_2) = \left( \dfrac{(1-m^2)x_1 + 2m(y_1-c)}{1+m^2}, \dfrac{2mx_1 - (1-m^2)y_1 + 2mc}{1+m^2} \right)$, where $(x_1, y_1)$ is the focus, $y = mx + c$ is the tangent line, and $(x_2, y_2)$ is the image of the focus.

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