Parabola
1.0 Conic Section
2.0 Parabola
3.0 Standard equation of Parabola
4.0 Focal distance of a point
5.0 General equation of Parabola
6.0 The generalized form of parabola: ${\left( {y - k} \right)^2} = 4a\left( {x - h} \right)$
7.0 Parametric Co-ordinates
7.1 Parametric relation between the coordinates of the ends of a focal chord of parabola
7.2 Important Results
8.0 Equation of tangent to a parabola
9.0 Point of intersection of tangents at any two points on the parabola
10.0 Equation of normal to the parabola
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
11.0 Relation between parametric coefficients if normal meets parabola
12.0 Important relations
13.0 Circle through co-normal points
14.0 Chord of contact
2.1 Important Terms
7.2 Important Results
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
1- Axis: The straight line passing through the focus and perpendicular to the directrix is called the axis of a conic section.
2- Vertex: The points of intersection of the conic section and the axis is(are) called vertex(vertices) of the conic section. It divides the line joining the focus and directrix in equal parts.
3- Focal chord: Any chord passing through focus is called focal chord.
4- Double ordinate: A straight line drawn perpendicular to the axis and terminated at both ends of the curve is a double ordinate of the conic section.
5- Latus rectum: The double ordinate passing through the focus is called the latus rectum of the conic.
6- Centre: The point which bisects every chord of the conic passing through it, is called centre of the conic section.
Question 1. Find the equation of parabola whose focus is $(-1,-2)$ and equation of directrix is $x - 2y + 3 = 0$.
Solution: The coordinates of focus is $S(-1,-2)$ and equation of directrix is $x - 2y + 3 = 0$. Let us assume the coordinates of point $P(h,k)$. From the definition of parabola, $$SP=PM$$
Therefore, $$\sqrt {{{\left( {h + 1} \right)}^2} + {{\left( {k + 2} \right)}^2}} = \left| {\frac{{h - 2k + 3}}{{\sqrt {{1^2} + {2^2}} }}} \right|$$ $$\sqrt {{{\left( {h + 1} \right)}^2} + {{\left( {k + 2} \right)}^2}} = \left| {\frac{{h - 2k + 3}}{{\sqrt 5 }}} \right|$$
Squaring both sides we get, $$5\left( {{h^2} + 1 + 2h + {k^2} + 4 + 4k} \right) = {h^2} + 4{k^2} - 4kh + 9 + 6h - 12k$$ $$4{h^2} + {k^2} + 4hk + 4h + 32k + 16 = 0$$
Therefore, the equation of parabola is $$4{x^2} + {y^2} + 4xy + 4x + 32y + 16 = 0$$
Question 2. Find the equation of parabola whose vertex is $(2,1)$ and equation of directrix is $x = y - 1$.
Solution: The coordinates of vertex is $V(2,1)$ and the equation of directrix is $x = y - 1$. Let us assume the
coordinates of focus $S\left( {\alpha ,\beta } \right)$. The slope of directrix is $1$. Since line $NV$ is perpendicular to the directrix, the slope of $NV = \frac{{ - 1}}{{slope\ of\ directrix}}$ i.e.,
Slope of $NV=-1$ and equation of line $NV$ passing through vertex $V(2,1)$ is $$y - 1 = - 1\left( {x - 2} \right)$$ $$x + y = 3$$
Solving it with equation of directrix we get the coordinates of point of intersection $N(1,2)$.
As we know that vertex divide the line joining focus and directrix in equal parts or in other words vertex $V$ is the mid-point of $N$ and $S$. Using mid-point formulae, we get $$2 = \frac{{\alpha + 1}}{2}\ and\ 1 = \frac{{\beta + 2}}{2}$$
or, $$\alpha = 3\ and\ \beta = 0$$
Let us assume the coordinates of point $P(h,k)$. From the definition of a parabola, $SP=PM$.
Therefore, $$\sqrt {{{\left( {h - 3} \right)}^2} + {{\left( k \right)}^2}} = \left| {\frac{{h - k + 1}}{{\sqrt {{1^2} + {1^2}} }}} \right|$$ $$\sqrt {{{\left( {h - 3} \right)}^2} + {{\left( k \right)}^2}} = \left| {\frac{{h - k + 1}}{{\sqrt 2 }}} \right|$$
Squaring both sides we get, $$2\left( {{h^2} - 9 - 6h + {k^2}} \right) = {h^2} + {k^2} + 1 + 2k - 2kh - 2h$$ $${h^2} + {k^2} + 2hk - 4h - 2k + 17 = 0$$
Therefore, the equation of parabola is $${x^2} + {y^2} + 2xy - 4x - 2y + 17 = 0$$
Question 3. The axis of parabola is along the line $y=x$ and distance of the vertex and focus from origin are $\sqrt 2 $ and $2\sqrt 2 $. respectively. If vertex and focus both lies in the first quadrant, find the equation of parabola.
Solution: As we know that the directrix is perpendicular to the axis of parabola, therefore, the slope of directrix is $-1$ and passing through the origin $(0,0)$. So,the equation of directrix is $$y=-x$$
As from the given data in the question, we can say that vertex is the midpoint of focus and origin. Let us assume the coordinates of focus $S(\beta ,\beta )$ and vertex $V(\alpha ,\alpha )$ as both lies on a line $y=x$. Now applying midpoint formulae, we get $$\alpha = \frac{{0 + \beta }}{2}$$
or, $$\beta = 2\alpha ...(1)$$
Now the distance of vertex $V(\alpha ,\alpha )$ from origin $(0,0)$ is $\sqrt 2 $.
Apply distance formulae between them we get, $$\sqrt 2 = \sqrt {{{\left( {\alpha - 0} \right)}^2} + {{\left( {\alpha - 0} \right)}^2}} $$
Squaring both sides, we get $$2 = 2{\alpha ^2}$$ or, $$\alpha = \pm 1$$
Therefore, from equation $(1)$, we get $$\beta = \pm 2$$
As given in question both vertex and focus lies in first quadrant, therefore the coordinates of vertex and focus are $V(1,1)$ and $S(2,2)$ respectively.
Therefore, $$\sqrt {{{\left( {h - 2} \right)}^2} + {{\left( {k - 2} \right)}^2}} = \left| {\frac{{h + k}}{{\sqrt {{1^2} + {1^2}} }}} \right|$$ $$\sqrt {{{\left( {h - 2} \right)}^2} + {{\left( {k - 2} \right)}^2}} = \left| {\frac{{h + k}}{{\sqrt 2 }}} \right|$$
Squaring both sides we get, $$2\left( {{h^2} + 4 - 4h + {k^2} + 4 - 4k} \right) = {h^2} + {k^2} + 2kh$$
$${h^2} + {k^2} - 2hk - 8h - 8k + 16 = 0$$
Therefore, the equation of parabola is ${x^2} + {y^2} - 2xy - 8x - 8y + 16 = 0$.
