The tangent at the point $(1, 2)$ to the curve $y^2 = 4x$ makes an angle $\theta$ with the positive direction of the $X$-axis. Then $\theta =$ (in $^{\circ}$)
- A$60$
- B$30$
- C$90$
- D$45$
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Similar Questions
Find the equation of the tangent to the parabola $x^2 = y$ at one of the endpoints of its latus rectum in the first quadrant.
Easy
View SolutionFor the parabola $y^2+6y-2x+5=0$,match the items in List-$I$ with the suitable item in List-$II$ given below:
List-$I$ List-$II$ $(I)$ Vertex $(A)$ $(-\frac{3}{2}, -3)$ $(II)$ Focus $(B)$ $(\frac{3}{2}, -3)$ $(III)$ Equation of the directrix $(C)$ $2x+5=0$ $(IV)$ Equation of the axis $(D)$ $2x+y+3=0$ $(E)$ $y+3=0$ $(F)$ $(-2, -3)$
| List-$I$ | List-$II$ |
| $(I)$ Vertex | $(A)$ $(-\frac{3}{2}, -3)$ |
| $(II)$ Focus | $(B)$ $(\frac{3}{2}, -3)$ |
| $(III)$ Equation of the directrix | $(C)$ $2x+5=0$ |
| $(IV)$ Equation of the axis | $(D)$ $2x+y+3=0$ |
| $(E)$ $y+3=0$ | |
| $(F)$ $(-2, -3)$ |
$A$ point on the parabola whose axis is parallel to the $X$-axis and which passes through the points $(0,1), (3,0), (0,-2)$ is
The locus of the middle points of all chords of the parabola $y^2 = 4ax$ passing through the vertex is
The points of intersection of the parabolas $y^2=5x$ and $x^2=5y$ lie on the line
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