The sum of diameters of the circles that touc | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The equation of the parabola is $x^2 = \frac{64}{15}(y - \frac{3}{5})$.The equation of the tangent at point $P\left(\frac{8}{5}, \frac{6}{5}\right

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) The equation of the parabola is $x^2 = \frac{64}{15}(y - \frac{3}{5})$.The equation of the tangent at point $P\left(\frac{8}{5}, \frac{6}{5}\right)$ is given by $x \cdot \frac{8}{5} = \frac{32}{15}(y + \frac{6}{5} - \frac{6}{5})$,which simplifies to $3x - 4y = 0$.The family of circles touching the parabola at $P$ is $(x - \frac{8}{5})^2 + (y - \frac{6}{5})^2 + \lambda(3x - 4y) = 0$.Expanding this,we get $x^2 + y^2 + x(3\lambda - \frac{16}{5}) + y(-4\lambda - \frac{12}{5}) + 4 = 0$.Since the circle touches the $y$-axis,the condition $f^2 = c$ must hold,where $f = \frac{1}{2}(-4\lambda - \frac{12}{5}) = -2\lambda - \frac{6}{5}$ and $c = 4$.Thus,$(-2\lambda - \frac{6}{5})^2 = 4$,which implies $|-2\lambda - \frac{6}{5}| = 2$.Case $1$: $-2\lambda - \frac{6}{5} = 2 \implies -2\lambda = \frac{16}{5} \implies \lambda = -\frac{8}{5}$. The radius $r_1 = \sqrt{g^2 + f^2 - c} = \sqrt{(\frac{3\lambda - 16/5}{2})^2 + 2^2 - 4} = |\frac{3\lambda - 16/5}{2}| = |\frac{3(-8/5) - 16/5}{2}| = |\frac{-40/5}{2}| = 4$. Diameter $d_1 = 8$.Case $2$: $-2\lambda - \frac{6}{5} = -2 \implies -2\lambda = -\frac{4}{5} \implies \lambda = \frac{2}{5}$. The radius $r_2 = |\frac{3(2/5) - 16/5}{2}| = |\frac{-10/5}{2}| = 1$. Diameter $d_2 = 2$.The sum of diameters is $d_1 + d_2 = 8 + 2 = 10$.

The sum of diameters of the circles that touch $(i)$ the parabola $75x^2 = 64(5y - 3)$ at the point $\left(\frac{8}{5}, \frac{6}{5}\right)$ and $(ii)$ the $y$-axis,is equal to $......$

Similar Questions

Let the parabola $y=x^2+px-3$ meet the coordinate axes at the points $P, Q$ and $R$. If the circle $C$ with center at $(-1,-1)$ passes through the points $P, Q$ and $R$,then the area of $\triangle PQR$ is:

The focal chord to $y^2 = 16x$ is tangent to $(x - 6)^2 + y^2 = 2$. Then,the possible values of the slope of this chord are:

The locus of the centre of the circle which touches the circle ${x^2} + {(y - 1)^2} = 1$ externally and also touches the $x$-axis is

$A$ focal chord to $y^2 = 16x$ is a tangent to $(x - 6)^2 + y^2 = 2$. Then the possible values of the slope of this chord are:

$A$ parabola $y = ax^2 + bx + c$ crosses the $x$-axis at $(\alpha, 0)$ and $(\beta, 0)$,both to the right of the origin. $A$ circle also passes through these two points. The length of a tangent from the origin to the circle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module