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Parabola Part 1

Section: Mathematics

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Question : 34 of 100

Marks: +1, -0

Let

x^2+y^2+Ax+By+C=0

be a circle passing through

(0,6)

and touching the parabola

y=x^2

(2,4)

A+C

[24-Jun-2022-Shift-1]

16
$\frac{88}{5}$
72
$-8$

Solution: 👈: Video Solution

For tangent to parabola

y=x^2

at

(2,4)

.\;\frac{dy}{dx}\Big|_{(2,4)}=4

Equation of tangent is

y-4=4(x-2)

\Rightarrow 4x-y-4=0

Family of circle can be given by

(x-2)^2+(y-4)^2+\lambda(4x-y-4)=0

As it passes through

(0,6)

2^2+2^2+\lambda(-10)=0

\Rightarrow \lambda=\frac{4}{5}

Equation of circle is

(x-2)^2+(y-4)^2+\frac{4}{5}(4x-y-4)=0

\Rightarrow(x^2+y^2-4x-8y+20)+\left(\frac{16}{5}x-\frac{4}{5}y-\frac{16}{5}\right)=0

A=-4+\frac{16}{5}, C=20-\frac{16}{5}

\;\text{So,}\; A+C=16

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