If the tangents PQ and PR are drawn from a variable point P to the ellipse x^ 2 16 + y^ 2 9 = 1, such that the fourth point S of the parallelogram PQSR lies on the circumcircle of the Δ PQR , then the area (in sq.

Correct answer: C. 25π Parallelogram PQSR is a cyclic quadrilateral ⇒ Parallelogram is a rectangle ⇒ ∠QPR = 90 ^ o Let, the equation of the tangent from P (h, k) to the ellipse is y = mx ± 16m^ 2 + 9 ⇒ (k - mh) ^ 2 = 16m^ 2 + 9 ∵ m _ 1 m _ 2 = - 1 ⇒ h^ 2 - 16 = - k^ 2 + 9 ⇒ locus of P is x^ 2 + y^ 2 = 25 Hence, the area of locus of P is 25π & sq. units

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Ellipse

NTA Abhyas 2020 · Mathematics · Ellipse

If the tangents PQ and PR are drawn from a variable point P to the...

If the tangents

PQ

and

PR

are drawn from a variable point

P

to the ellipse

\dfrac{x^{2}}{16} + \dfrac{y^{2}}{9} = 1,

such that the fourth point

S

of the parallelogram

PQSR

lies on the circumcircle of the

\Delta PQR

, then the area (in sq. units) of the locus of

P

Options

A $7\pi$
B $16\pi$
C $25\pi$
D $9\pi$

Correct answer

C. $25\pi$

Step-by-step solution

Parallelogram $PQSR$ is a cyclic quadrilateral
$⇒$ Parallelogram is a rectangle
$⇒ ∠QPR = {90}^{o}$
Let, the equation of the tangent from $P \left(h, k\right) to the ellipse is$ y = mx \pm \sqrt{16m^{2} + 9} $⇒ {\left(k - mh\right)}$ ^{2} = 16m^{2} + 9 $ $ ∵ {m}_{1}{m}_{2} = - 1 ⇒ h^{2} - 16 = - k^{2} + 9 $ $ ⇒ $locus of$ P $is$ x^{2} + y^{2} = 25 $Hence, the area of locus of$ P $is$ 25\pi $& sq. units$

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