What is the equation of the tangent to the parabola $y^2 = 12x$ at the upper end of its latus rectum?
- A$y = x + 3$
- B$x + y = 3$
- C$y = x - 3$
- DNone of these
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Similar Questions
Find the equation of the normal to the parabola $y^2 + 12x = 0$ at the upper end of its latus rectum.
Medium
View SolutionThe length of the latus rectum of the parabola $169\{(x-1)^2+(y-3)^2\}=(5x-12y+17)^2$ is
Consider the parabola $y^2+2x+2y-3=0$ and match the items of List-$I$ with those of the List-$II$.
$A. \ 2x-5=0$ $I. \ \text{Vertex}$ $B. \ (\frac{3}{2}, -1)$ $II. \ \text{Focus}$ $C. \ y+1=0$ $III. \ \text{Equation of directrix}$ $D. \ (2, -1)$ $IV. \ \text{Equation of the axis}$ $V. \ \text{Equation of the Latus rectum}$
The correct match is:
| $A. \ 2x-5=0$ | $I. \ \text{Vertex}$ |
| $B. \ (\frac{3}{2}, -1)$ | $II. \ \text{Focus}$ |
| $C. \ y+1=0$ | $III. \ \text{Equation of directrix}$ |
| $D. \ (2, -1)$ | $IV. \ \text{Equation of the axis}$ |
| $V. \ \text{Equation of the Latus rectum}$ |
The correct match is:
The focal distance of a point $(5, 5)$ on the parabola $x^2 - 2x - 4y + 5 = 0$ is
What is the angle in degrees between the tangents drawn from the origin to the parabola $y^2 = 4a(x - a)$?
Difficult
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