If the tangent drawn at the point P(3 sqrt{2} | vedclass

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

Question

(C) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{16} = 1$. Here $a^2 = 9$ and $b^2 = 16$,so $a = 3$ and $b = 4$.The equation of the ta

Vedclass · Accepted Answer

$\frac{12 \sqrt{2}-20}{5}$

Vedclass · Answer

(C) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{16} = 1$. Here $a^2 = 9$ and $b^2 = 16$,so $a = 3$ and $b = 4$.The equation of the tangent at $P(x_1, y_1) = (3 \sqrt{2}, 4)$ is given by $\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$.Substituting the values,we get $\frac{x(3 \sqrt{2})}{9} - \frac{y(4)}{16} = 1$,which simplifies to $\frac{x \sqrt{2}}{3} - \frac{y}{4} = 1$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$.The directrix of the hyperbola is $x = \frac{a}{e} = \frac{3}{5/3} = \frac{9}{5}$.Since the point $Q(\alpha, \beta)$ lies on the directrix,$\alpha = \frac{9}{5}$.Substituting $x = \frac{9}{5}$ into the tangent equation: $\frac{(9/5) \sqrt{2}}{3} - \frac{y}{4} = 1$.$\frac{3 \sqrt{2}}{5} - \frac{y}{4} = 1 \implies \frac{y}{4} = \frac{3 \sqrt{2}}{5} - 1 = \frac{3 \sqrt{2} - 5}{5}$.Thus,$y = \beta = \frac{12 \sqrt{2} - 20}{5}$.Since $Q$ is in the fourth quadrant,we check the sign. $\sqrt{2} \approx 1.414$,so $12(1.414) \approx 16.968$. $16.968 - 20 < 0$,which is consistent with the fourth quadrant.

If the tangent drawn at the point $P(3 \sqrt{2}, 4)$ on the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ meets its directrix at $Q(\alpha, \beta)$ in the fourth quadrant,then $\beta=$

Similar Questions

If for a hyperbola the ratio of the length of the conjugate axis to the length of the transverse axis is $3:2$,then the ratio of the distance between the foci to the distance between the two directrices is

If the equation $\frac{x^2}{7 - k} + \frac{y^2}{5 - k} = 1$ represents a hyperbola,then:

The equation of the conic with focus at $(1, -1)$,directrix along $x - y + 1 = 0$ and with eccentricity $e = \sqrt{2}$ is:

The locus of the point of intersection of lines $(x + y)t = a$ and $x - y = at$,where $t$ is the parameter,is

If the area of the region bounded by $16x^2 - 9y^2 = 144$ and $8x - 3y = 24$ is $A$, then $3(A + 6 \ln(3))$ is equal to . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module