Let equation of two diameters of a circle x ^{2}+ y ^{2} -2 x +2 fy +1=0 be 2 px

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10-2. Parabola, Ellipse, Hyperbola

normal

Let the equation of two diameters of a circle $x ^{2}+ y ^{2}$ $-2 x +2 fy +1=0$ be $2 px - y =1$ and $2 x + py =4 p$. Then the slope $m \in(0, \infty)$ of the tangent to the hyperbola $3 x^{2}-y^{2}=3$ passing through the centre of the circle is equal to $......$

$6$

$2$

$4$

$8$

(JEE MAIN-2022)

Solution

$2 p+f-1=0$

$2-p f-4 p=0$

$2=p(f+4)$

$p=\frac{2}{f+4}$

$2 p=1-f$

$\frac{4}{f+4}=1-f$

$f^{2}+3 f=0$

$f=0 \text { or }-3$

Hyperbola $3 x ^{2}- y ^{2}=3, x ^{2}-\frac{ y ^{2}}{3}=1$

$y=m x \pm \sqrt{m^{2}-3}$

It passes $(1,0)$

$o=m \pm \sqrt{m^{2}-3}$

$m$ tends $\infty$

$\text { It passes }(1,3)$

$3=m \pm \sqrt{m^{2}-3}$

$(3-m)^{2}=m^{2}-3$

$m=2$

Standard 11

Mathematics

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Solution

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If $\frac{{{{\left( {3x – 4y – z} \right)}^2}}}{{100}} – {\frac{{\left( {4x + 3y – 1} \right)}}{{225}}^2} = 1$ then
length of latusrectum of hyperbola is