A line passes through the point (-1, 1) and m | vedclass

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

Question

(B) Given the angle $	heta = \sin^{-1}(\frac{3}{5})$, we have $	an 	heta = \frac{3}{4}$. The slope of the line is $m = \frac{3}{4}$.The equation of

Vedclass · Accepted Answer

$\frac{5}{4}$ unit

Vedclass · Answer

(B) Given the angle $	heta = \sin^{-1}(\frac{3}{5})$, we have $	an 	heta = \frac{3}{4}$. The slope of the line is $m = \frac{3}{4}$.The equation of the line passing through $(-1, 1)$ with slope $m = \frac{3}{4}$ is $y - 1 = \frac{3}{4}(x + 1)$, which simplifies to $4y - 4 = 3x + 3$, or $4y = 3x + 7$.Substitute $4y = 3x + 7$ into the curve equation $x^2 = 4y - 9$:$x^2 = (3x + 7) - 9$$x^2 - 3x + 2 = 0$$(x - 1)(x - 2) = 0$So, $x = 1$ and $x = 2$.For $x = 1$, $4y = 3(1) + 7 = 10 \Rightarrow y = \frac{5}{2}$. Point $A = (1, \frac{5}{2})$.For $x = 2$, $4y = 3(2) + 7 = 13 \Rightarrow y = \frac{13}{4}$. Point $B = (2, \frac{13}{4})$.The distance $|AB| = \sqrt{(2 - 1)^2 + (\frac{13}{4} - \frac{5}{2})^2} = \sqrt{1^2 + (\frac{13-10}{4})^2} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$ units.

$A$ line passes through the point $(-1, 1)$ and makes an angle $\sin^{-1}(\frac{3}{5})$ in the positive direction of the $x$-axis. If this line meets the curve $x^2 = 4y - 9$ at points $A$ and $B$, then the length $|AB|$ is equal to

Similar Questions

The parametric equations of the parabola $x^2-8 x+12 y+15=0$ are

From a point $(C, 0)$,three normals are drawn to the parabola $y^2=x$. Then,

If $(0, 6)$ and $(0, 3)$ are respectively the vertex and focus of a parabola,then its equation is

The Cartesian coordinates of the point on the parabola $y^2 = -16x$,whose parameter is $t = \frac{1}{2}$,are

If tangents are drawn from the point $(-1, 2)$ to the parabola $y^2 = 4x$, what is the area of the triangle formed by the chord of contact and the tangents (in $\sqrt{2}$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module