Question

The chord joining two points ${\theta _1}{\text{ and }}{\theta _2}$on the ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ such that $\tan {\theta _1}\tan {\theta _2} = - \dfrac{{{a^2}}}{{{b^2}}}$ will subtend a right angle at
$(a){\text{ focus}} \\\ (b){\text{ center}} \\\ (c){\text{ end of the major axis}} \\\ (d){\text{ end of the minor axis}} \\\$

Answer 1

Hint: In this question suppose two points ${\theta _1}$ and ${\theta _2}$ such that ${\theta _1} = \left( {a\cos {\theta _1},b\sin {\theta _1}} \right)$ and ${\theta _2} = \left( {a\cos {\theta _2},b\sin {\theta _2}} \right)$through which the chord passes. Then use the concept of slope of line passing through two given points to find the slope of $O{\theta _1}{\text{ and O}}{\theta _2}$ where O is the origin. Use the concept that if two lines are perpendicular then their slopes are related as ${m_1} \times {m_2} = - 1$.

Complete step-by-step answer:

The chord joining two points $\left( {{\theta _1},{\theta _2}} \right)$ on the ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ is shown above.
As we know that the ellipse is having a center (O) = (0, 0) is also shown in the figure.
Let us suppose the point ${\theta _1} = \left( {a\cos {\theta _1},b\sin {\theta _1}} \right)$ and ${\theta _2} = \left( {a\cos {\theta _2},b\sin {\theta _2}} \right)$ is also shown in the figure.
Now as we know that the slope between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given as
Slope (m) = $\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
So find out the slopes of $\left( {O{\theta _1}} \right)$ and $\left( {O{\theta _2}} \right)$.
Let O = $(x_1, y_1)$ = (0, 0)
${\theta _1} = \left( {{x_2},{y_2}} \right) = \left( {a\cos {\theta _1},b\sin {\theta _1}} \right)$
${\theta _2} = \left( {{x_3},{y_3}} \right) = \left( {a\cos {\theta _2},b\sin {\theta _2}} \right)$
So let the slope of $\left( {0{\theta _1}} \right)$ be m1.
$\Rightarrow {m_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \dfrac{{b\sin {\theta _1} - 0}}{{a\cos {\theta _1} - 0}} = \dfrac{b}{a}\tan {\theta _1}$
Now let the slope of $\left( {O{\theta _2}} \right)$ be m2.
$\Rightarrow {m_1} = \dfrac{{{y_3} - {y_1}}}{{{x_3} - {x_1}}} = \dfrac{{b\sin {\theta _2} - 0}}{{a\cos {\theta _2} - 0}} = \dfrac{b}{a}\tan {\theta _2}$
Now multiply the slopes we have
$\Rightarrow {m_1} \times {m_2} = \dfrac{b}{a}\tan {\theta _1} \times \dfrac{b}{a}\tan {\theta _2} = \dfrac{{{b^2}}}{{{a^2}}}\tan {\theta _1}\tan {\theta _2}$........................ (1)
Now it is given that
$\tan {\theta _1}\tan {\theta _2} = - \dfrac{{{a^2}}}{{{b^2}}}$
Now substitute this value in equation (1) we have,
$\Rightarrow {m_1} \times {m_2} = \dfrac{{{b^2}}}{{{a^2}}}\tan {\theta _1}\tan {\theta _2} = \dfrac{{{b^2}}}{{{a^2}}} \times \dfrac{{ - {a^2}}}{{{b^2}}} = - 1$
So multiplication of slopes is (-1) which is the condition of the right angle.
Therefore the chord joining two points $\left( {{\theta _1},{\theta _2}} \right)$ on the ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ will subtend a right angle at origin or center.
Hence option (B) is correct.

Note: The center of the given ellipse that is $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ is (0, 0) that is the origin that’s why option (c) is correct. The equation of shifted ellipse or the ellipse whose center is not at origin is given by $\dfrac{{{{\left( {x - p} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - q} \right)}^2}}}{{{b^2}}} = 1$ here the center is at (p, q).