Find the differential equation of the family of parabolas with vertex at (h, 0) and the principal axis along the x-axis.
Answer
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Hint: Here, we need to find the differential equation i.e.., an equation in the form of $\frac{{dy}}{{dx}}$by using the property of the parabola i.e..,$PS = PM$.
Given, vertex at (h, 0) and principal axis along x-axis.
Let $P(x,y)$be a point on the parabola ${y^2} = 4ax$
As we know that the focus ‘S’ coordinates will be$(a + h,0)$. Therefore, $PS = \sqrt {({{(x - (a + h))}^2} + {y^2}} $.
As we know that the perpendicular distance from point $({x_1},{y_1})$ to a line $ax + by + c = 0$is$\frac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$. It is also given that the principal axis is along the x-axis .Therefore, PM can be written as the Perpendicular distance from Point p to the principal axis.
Hence $PM = \frac{{\left| {x + a - h} \right|}}{{\sqrt {{1^2} + 0} }} = \left| {x + a - h} \right|$
From Parabola, we know that $PS = PM$
Hence, substituting the values of PS and PM we get
$\sqrt {({{(x - (a + h))}^2} + {y^2}} = \left| {x + a - h} \right|$
Squaring on both sides we get
\[
\Rightarrow {(x - {(a + h)^2})^2} + {y^2} = {(x + (a - h))^2} \\
\Rightarrow {x^2} + {(a + h)^2} - 2x(a + h) + {y^2} = {x^2} + {(a - h)^2} + 2x(a - h) \\
\Rightarrow {x^2} + {a^2} + {h^2} + 2ah - 2ax - 2ah + {y^2} = {x^2} + {a^2} + {h^2} - 2ah + 2ax - 2ah \\
\Rightarrow 4ah - 4ax + {y^2} = 0 \\
\Rightarrow {y^2} = 4a(x - h) \to (1) \\
\]
Differentiating equation (1) with respect to x we get
$2y\frac{{dy}}{{dx}} = 4a(1 - 0)$
$2y\frac{{dy}}{{dx}} = 4a \to (2)$
Putting the value of 4a in equation (1) we get
$
\Rightarrow {y^2} = 2y\frac{{dy}}{{dx}}(x - h) \\
\Rightarrow \frac{{dy}}{{dx}} = \frac{y}{{2(x - h)}} \\
$
Hence the differential equation the differential equation of the family of parabolas with vertex at (h, 0) and the principal axis along the x-axis is$\frac{y}{{2(x - h)}}$.
Note: To solve the given problem, concepts of the parabola have to be known properly and use the formulae$PS = PM$.
Given, vertex at (h, 0) and principal axis along x-axis.
Let $P(x,y)$be a point on the parabola ${y^2} = 4ax$
As we know that the focus ‘S’ coordinates will be$(a + h,0)$. Therefore, $PS = \sqrt {({{(x - (a + h))}^2} + {y^2}} $.
As we know that the perpendicular distance from point $({x_1},{y_1})$ to a line $ax + by + c = 0$is$\frac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$. It is also given that the principal axis is along the x-axis .Therefore, PM can be written as the Perpendicular distance from Point p to the principal axis.
Hence $PM = \frac{{\left| {x + a - h} \right|}}{{\sqrt {{1^2} + 0} }} = \left| {x + a - h} \right|$
From Parabola, we know that $PS = PM$
Hence, substituting the values of PS and PM we get
$\sqrt {({{(x - (a + h))}^2} + {y^2}} = \left| {x + a - h} \right|$
Squaring on both sides we get
\[
\Rightarrow {(x - {(a + h)^2})^2} + {y^2} = {(x + (a - h))^2} \\
\Rightarrow {x^2} + {(a + h)^2} - 2x(a + h) + {y^2} = {x^2} + {(a - h)^2} + 2x(a - h) \\
\Rightarrow {x^2} + {a^2} + {h^2} + 2ah - 2ax - 2ah + {y^2} = {x^2} + {a^2} + {h^2} - 2ah + 2ax - 2ah \\
\Rightarrow 4ah - 4ax + {y^2} = 0 \\
\Rightarrow {y^2} = 4a(x - h) \to (1) \\
\]
Differentiating equation (1) with respect to x we get
$2y\frac{{dy}}{{dx}} = 4a(1 - 0)$
$2y\frac{{dy}}{{dx}} = 4a \to (2)$
Putting the value of 4a in equation (1) we get
$
\Rightarrow {y^2} = 2y\frac{{dy}}{{dx}}(x - h) \\
\Rightarrow \frac{{dy}}{{dx}} = \frac{y}{{2(x - h)}} \\
$
Hence the differential equation the differential equation of the family of parabolas with vertex at (h, 0) and the principal axis along the x-axis is$\frac{y}{{2(x - h)}}$.
Note: To solve the given problem, concepts of the parabola have to be known properly and use the formulae$PS = PM$.
Last updated date: 21st Sep 2023
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