8.1 Tangent at a given point

1.1 General Equation of Conic Section

2.1 Important Terms

7.1 Parametric relation between the coordinates of the ends of a focal chord of parabola
7.2 Important Results

8.1 Tangent at a given point
8.2 Tangent in terms of slope
8.3 Parametric form

9.1 Important Result

10.1 Point form
10.2 Slope form
10.3 Parametric form
10.4 To find the number of normal drawn from a point to a parabola (CONCEPT THROUGH QUESTIONS):
10.5 Point of intersection of normal at any two points on the parabola

13.1 Pair of tangents

14.1 Length of chord of contact (QR)
14.2 Equation of chord of parabola whose midpoint is given

If a point $P({x_1},{y_1})$ lies on a parabola, use $T=0$ to find the equation of tangent at point $P$.

$y{y_1} = 2a\left( {x + {x_1}} \right)$ $x{x_1} = 2a\left( {y + {y_1}} \right)$

Or, We can use differentiation to find the slope of tangent at point $P$ and then find the equation of tangent. Let us assume the equation of parabola be $${y^2} = 4ax...(1)$$

Differentiate equation $(1)$ with respect to $y$, we get $$2y\frac{{dy}}{{dx}} = 4a$$ $$\frac{{dy}}{{dx}} = \frac{{2a}}{y}$$

Slope of tangent at $P({x_1},{y_1})$ is $${\frac{{dy}}{{dx}}_{({x_1},{y_1})}} = \frac{{2a}}{{{y_1}}}$$

Therefore, equation of tangent is $$y - {y_1} = \frac{{dy}}{{dx}}\left( {x - {x_1}} \right)$$ $$y - {y_1} = \frac{{2a}}{{{y_1}}}\left( {x - {x_1}} \right)$$

Question 6. If a tangent is drawn at any point on the parabola ${y^2} = 4ax$ which cuts coordinate axis at $A$ and $B$. Find locus of mid-point of $AB$.

Solution: Let us assume that the tangent is drawn to the parabola at a point $P(a{t^2},2at)$ which cuts the coordinate axis at points $A(0,\beta )$ and $B(\alpha ,0)$ as shown in figure $18$.

Using $T=0$, the equation of tangent at point $P(a{t^2},2at)$ is $$2aty = 2a(x + a{t^2})$$ $$y = \frac{x}{t} + at...(1)$$

On comparing the equation $(1)$ with $y=mx+c$, we get the slope of tangent i.e., $$m = \frac{1}{t}...(2)$$

Since $A$ and $B$ lies on the tangent, slope of tangent using two given points is $$m = \frac{{0 - \beta }}{{\alpha - 0}} = - \frac{\beta }{\alpha }...(3)$$

Equations $(2)$ and $(3)$ are same, therefore, $$ - \frac{\beta }{\alpha } = \frac{1}{t}...(4)$$

Now, The mid-point of $AB$ is $M(h,k)$, applying mid-point formulae, we get $$h = \frac{\alpha }{2}$$

or, $$\alpha = 2h$$ and $$k = \frac{\beta }{2}$$ or, $$\beta = 2k$$

Put the values of $\alpha $ and $\beta $ in equation $(4)$, we get $$ - \frac{{2k}}{{2h}} = \frac{1}{t}$$

or, $$h+kt=0$$

Therefore, the locus of mid-point of $AB$ is a straight line having equation $$x+ty=0$$