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Definite Integrals

Let d1,d2 be ...

Question

Let $d_{1}, d_{2}$ be the longest and shortest distance of a point $P (7, 2)$ from the curve $x^{2} + y^{2} - 10 x - 14 y - 51 = 0.$ Then the geometric mean of $d_{1}$ and $d_{2}$ is equal to

A

\sqrt{6}

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B

2 \sqrt{6}

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C

3 \sqrt{6}

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D

4 \sqrt{6}

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Solution

The correct option is D $4 \sqrt{6}$
$x^{2} + y^{2} - 10 x - 14 y - 51 = 0$ is a circle centred at $C (5, 7)$ with radius $5 \sqrt{5} .$

$S_{1} = 7^{2} + 2^{2} - 70 - 28 - 51 = - 96 < 0$
$∴ P$ lies inside the circle.

Hence, $d_{1} = P R = r + C P$ and $d_{2} = P Q = r - C P$
$C P = \sqrt{(7 - 5)^{2} + (2 - 7)^{2}} = \sqrt{29}$

Geometric mean of $d_{1}, d_{2}$ is
$\sqrt{d_{1} d_{2}}$
$= \sqrt{r^{2} - (C P)^{2}}$
$= \sqrt{125 - 29}$
$= 4 \sqrt{6}$

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Similar questions

d_{1}

d_{2}

are the longest and shortest distances of

(- 7, 2)

from the curve whose equation is

x^{2} + y^{2} - 10 x - 14 y = 51

then Geometric mean of

d_{1}

d_{2}

d_{1}

d_{2}

are the longest and the shortest distances of the point

P (- 7, 2)

from the circle

x^{2} + y^{2} - 10 x - 14 y - 51 = 0,

then the value of

d_{1}^{2} + d_{2}^{2}

p

q

be the longest distance and the shortest distance respectively of the point

(- 7, 2)

(α, β)

on the curve whose equation is

x^{2} + y^{2} - 10 x - 14 y - 51 = 0

p

q

p

q

be the longest distance and the shortest distance respectively of the point

(- 7, 2)

(α, β)

on the curve whose equation is

x^{2} + y^{2} - 10 x - 14 y - 51 = 0

G M

p

q

D_{1} = ∣ ∣ ∣ ∣ \begin{matrix} x & a & b - 1 & 0 & x x & 2 & 1 \end{matrix} ∣ ∣ ∣ ∣

D_{2} = ∣ ∣ ∣ ∣ \begin{matrix} c x^{2} & 2 a & - b x & 2 & 1 - 1 & 0 & x \end{matrix} ∣ ∣ ∣ ∣ .

If all the roots of

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