The point of contact of the tangent to the pa | vedclass

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(A) The equation of the tangent to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $yt = x + at^2$.Here,$4a = 9$,so $a = \frac{9}{4}$.The tangent p

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$\left(\frac{4}{9}, 2\right)$

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(A) The equation of the tangent to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $yt = x + at^2$.Here,$4a = 9$,so $a = \frac{9}{4}$.The tangent passes through $(4, 10)$,so $10t = 4 + \frac{9}{4}t^2$.Multiplying by $4$,we get $40t = 16 + 9t^2$,or $9t^2 - 40t + 16 = 0$.Factoring the quadratic: $(9t - 4)(t - 4) = 0$,which gives $t = 4$ or $t = \frac{4}{9}$.The slope of the tangent is $m = \frac{1}{t} = 	an 	heta$.Given $	an 	heta > 2$,we have $\frac{1}{t} > 2$,which implies $t Thus,$t = \frac{4}{9}$ is the valid parameter.The point of contact is $(at^2, 2at) = \left(\frac{9}{4} 	imes \left(\frac{4}{9}ight)^2, 2 	imes \frac{9}{4} 	imes \frac{4}{9}ight) = \left(\frac{4}{9}, 2ight)$.

The point of contact of the tangent to the parabola $y^2=9x$ which passes through the point $(4, 10)$ and makes an angle $\theta$ with the positive side of the axis of the parabola where $\tan \theta > 2$,is

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