Question

Statement-1: Feet of the perpendicular drawn from the foot of an ellipse $4{x^2} + {y^2} = 16$ on the line $2\sqrt 3 x + y = 8$ lie on the circle ${x^2} + {y^2} = 16$.
Statement-2: If a perpendicular is drawn from foci of an ellipse to its any tangent then feet of these perpendicular lie on the direct circle of the ellipse.
A) Statement-1 is True, Statement-2 is True: Statement-2 is a correct explanation for Statement-1.
B) Statement-1 is True, Statement-2 is True: Statement-2 is not a correct explanation for Statement-1.
C) Statement-1 is True, Statement-2 is False
D) Statement-1 is False, Statement-2 is True.

Answer 1

First write the equation of an ellipse in standard form, it shows the ellipse is vertical. Now, find the foci of the ellipse. Then find the equation of the line perpendicular to the tangent and pass through the focus and square it. Now square the equation of the tangent and add with the equation of line perpendicular to it. The equation of the circle will be obtained.

Complete step-by-step answer:
Given: - The equation of the ellipse is $4{x^2} + {y^2} = 16$.
The equation of the line is $2\sqrt 3 x + y = 8$.
The equation of the circle is ${x^2} + {y^2} = 16$
Write the equation of an ellipse in standard form.
Divide both side of the equation by 16,
$\Rightarrow$ $\dfrac{{4{x^2}}}{{16}} + \dfrac{{{y^2}}}{{16}} = \dfrac{{16}}{{16}}$
Cancel out the common factor from the numerator and denominator from both sides,
$\Rightarrow$ $\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{{16}} = 1$
Since $b > a$. Then, it is a vertical ellipse.
The coordinates of the focus of the ellipse will be,
$\Rightarrow$ $\left( {0, \pm ae} \right)$
Substitute the value to find the focus,
$\Rightarrow$ $\left( {0, \pm ae} \right) = \left( {0, \pm 4\sqrt {1 - \dfrac{4}{{16}}} } \right)$
Take LCM inside the root,
$\Rightarrow$ $\left( {0, \pm ae} \right) = \left( {0, \pm 4\sqrt {\dfrac{{16 - 4}}{{16}}} } \right)$
Subtract 4 from 16 and cancel out the common factor from the numerator and denominator,
$\Rightarrow$ $\left( {0, \pm ae} \right) = \left( {0, \pm 4\sqrt {\dfrac{3}{4}} } \right)$
Apply square root on the denominator and cancel out the common factor from the numerator and denominator,
$\Rightarrow$ $\left( {0, \pm ae} \right) = \left( {0, \pm 2\sqrt 3 } \right)$
Equation of the line perpendicular to the tangent and passes through focus is,
$\Rightarrow$ $y - 2\sqrt 3 = - \dfrac{1}{{ - 2\sqrt 3 }}\left( {x - 0} \right)$
Cancel out the negative sign and cross multiply the terms,
$\Rightarrow$ $2\sqrt 3 y - 12 = x$
Move the variable on the left side and constant on the right side,
$\Rightarrow$ $2\sqrt 3 y - x = 12$
Square on both sides of the equation,
$\Rightarrow$ ${\left( {2\sqrt 3 y - x} \right)^2} = {\left( {12} \right)^2}$
Open the brackets and square the terms,
$\Rightarrow$ $12{y^2} + {x^2} - 4\sqrt 3 xy = 144$.............…..(1)
The equation of the line is,
$2\sqrt 3 x + y = 8$
Square on both sides of the equation,
$\Rightarrow$ ${\left( {2\sqrt 3 x + y} \right)^2} = {\left( 8 \right)^2}$
Open the brackets and square the terms,
$\Rightarrow$ $12{x^2} + {y^2} + 4\sqrt 3 xy = 64$............…..(2)
Add both equations (1) and (2),
$\Rightarrow$ $13{x^2} + 13{y^2} = 208$
Divide both sides by 13,
$\Rightarrow$ $\dfrac{{13{x^2}}}{{13}} + \dfrac{{13{y^2}}}{{13}} = \dfrac{{208}}{{13}}$
Cancel out the common factors from the numerator and denominator from both sides,
$\Rightarrow$ ${x^2} + {y^2} = 16$
Thus, the locus of the feet of the perpendicular drawn from foci to any tangent of the ellipse is the auxiliary circle.
Hence, statement-1 is true and statement-2 is false.

Option C is the correct answer.

Note: The students might make mistakes while finding the focus of the ellipse $b > a$. The foci will lie on the y-axis.
An ellipse is a shape that looks like an oval or a flattened circle.
In geometry, an ellipse is a plane curve that results from the intersection of a cone by a plane in a way that produces a closed curve.
The equation of the ellipse is: $\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}}$
where, $\left( {h,k} \right)$ is the center of the ellipse.