If the line 3x - my + 5 = 0 is a tangent to t | vedclass

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

Question

(D) Given the hyperbola equation: $3x^2 - 4y^2 = 300$. Dividing by $300$,we get: $\frac{x^2}{100} - \frac{y^2}{75} = 1$. Here,$a^2 = 100$ and $b^2 = 7

Vedclass · Accepted Answer

$\frac{15}{7}$

Vedclass · Answer

(D) Given the hyperbola equation: $3x^2 - 4y^2 = 300$. Dividing by $300$,we get: $\frac{x^2}{100} - \frac{y^2}{75} = 1$. Here,$a^2 = 100$ and $b^2 = 75$. The line is $3x - my + 5 = 0$,which can be written as $my = 3x + 5$,or $y = \frac{3}{m}x + \frac{5}{m}$. Comparing with $y = Mx + c$,we have $M = \frac{3}{m}$ and $c = \frac{5}{m}$. The condition for the line $y = Mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2M^2 - b^2$. Substituting the values: $(\frac{5}{m})^2 = 100(\frac{3}{m})^2 - 75$. $\frac{25}{m^2} = \frac{900}{m^2} - 75$. $75 = \frac{900 - 25}{m^2} = \frac{875}{m^2}$. $m^2 = \frac{875}{75} = \frac{35}{3}$. The $Y$-intercept is $c = \frac{5}{m}$,so the square of the $Y$-intercept is $c^2 = \frac{25}{m^2}$. Substituting $m^2 = \frac{35}{3}$: $c^2 = 25 	imes \frac{3}{35} = \frac{5 	imes 3}{7} = \frac{15}{7}$.

If the line $3x - my + 5 = 0$ is a tangent to the hyperbola $3x^2 - 4y^2 = 300$,then the square of the $Y$-intercept made by this tangent line is:

Similar Questions

Two points $A$ and $B$ have coordinates $(1, 0)$ and $(-1, 0)$ respectively,and $Q$ is a point which satisfies the relation $AQ - BQ = \pm 1$. The locus of $Q$ is:

The product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes is

If $e_1$ and $e_2$ are respectively the eccentricities of the curves $9x^2 - 16y^2 - 144 = 0$ and $9x^2 - 16y^2 + 144 = 0$,then find the value of $\frac{e_1^2 e_2^2}{e_1^2 + e_2^2}$.

The equation of the hyperbola whose eccentricity is $\frac{5}{3}$ and the distance between the foci is $10$ units is:

If the latus rectum of a hyperbola through one focus subtends an angle of $60^{\circ}$ at the other focus,then its eccentricity is

Vedclass Test Series

Exam Paper Generator

Online Exam Module