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Question: Write the equation for the parabola and draw the graph. Vertex \(\left( -7,4 \right)\), axis of sy...

Write the equation for the parabola and draw the graph.
Vertex (7,4)\left( -7,4 \right), axis of symmetry x=7x=-7, measure of latus rectum 6,a<06,a< 0.

Explanation

Solution

Assume the equation of the parabola as y=ax2+bx+cy=a{{x}^{2}}+bx+c. Substitute the point (7,4)\left( -7,4 \right) in the parabola to form the first relation in a, b and c. Now, differentiate the equation of parabola and at the point (7,4)\left( -7,4 \right) substitute dydx=0\dfrac{dy}{dx}=0 to form a second relation. Convert y=ax2+bx+cy=a{{x}^{2}}+bx+c in the form 1a(y+D4a)=(x+b2a)2\dfrac{1}{a}\left( y+\dfrac{D}{4a} \right)={{\left( x+\dfrac{b}{2a} \right)}^{2}} and compare it with x2=4Ay{{x}^{2}}=4Ay. Use the formula : L=4AL=\left| 4A \right| to determine the length of latus rectum. Find the value of a using this and use the other two relations to find the value of b and c. Finally substitute them in y=ax2+bx+cy=a{{x}^{2}}+bx+c to get the answer.

Complete step-by-step solution:
We have been given that the axis of symmetry of parabola is x=7x=-7. Let us assume the equation of parabola as y=ax2+bx+cy=a{{x}^{2}}+bx+c. It is given that a<0a<0, therefore the parabola will be opening downwards.

Now the vertex (7,4)\left( -7,4 \right) lies on the parabola, therefore substituting its value in the equation, we get,
4=49a7b+c(i)\Rightarrow 4=49a-7b+c\ldots \ldots \ldots \left( i \right)
Also, we can clearly see that at point (7,4)\left( -7,4 \right) the slope of tangent (dydx)\left( \dfrac{dy}{dx} \right) to the parabola will be zero as it will be parallel to x-axis. So, on differentiating the function, we get,
dydx=2ax+b\Rightarrow \dfrac{dy}{dx}=2ax+b
Substituting dydx=0\dfrac{dy}{dx}=0 at (7,4)\left( -7,4 \right), we get,
0=14a+b 14a=b(ii) \begin{aligned} & \Rightarrow 0=-14a+b \\\ & \Rightarrow 14a=b\ldots \ldots \ldots \left( ii \right) \\\ \end{aligned}
Now, we know that y=ax2+bx+cy=a{{x}^{2}}+bx+c can be written in the whole square form by using the square method. So, we have,
y=a[(x+b2a)2D4a2]y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]
Here, D=b24acD={{b}^{2}}-4ac = discriminant of the quadratic equation.
ya+D4a2=(x+b2a)2 (x+b2a)2=1a(y+D4a) \begin{aligned} & \Rightarrow \dfrac{y}{a}+\dfrac{D}{4{{a}^{2}}}={{\left( x+\dfrac{b}{2a} \right)}^{2}} \\\ & \Rightarrow {{\left( x+\dfrac{b}{2a} \right)}^{2}}=\dfrac{1}{a}\left( y+\dfrac{D}{4a} \right) \\\ \end{aligned}
This can be written as X2=4mY{{X}^{2}}=4mY, where X=(x+b2a),Y=(y+D4a)X=\left( x+\dfrac{b}{2a} \right),Y=\left( y+\dfrac{D}{4a} \right) and 4m=1a4m=\dfrac{1}{a}. So, on comparing the above relation with the general equation of parabola given as x2=4Ay{{x}^{2}}=4Ay, we get,
Length of latus rectum = L=4AL=\left| 4A \right|
So, for X2=4mY{{X}^{2}}=4mY, latus rectum will be :
L=4m=1aL=\left| 4m \right|=\left| \dfrac{1}{a} \right|
It is given that measure of latus rectum is 6 and a<0a<0, therefore removing the modulus sign, we get,
6=1a a=16 \begin{aligned} & \Rightarrow 6=\dfrac{-1}{a} \\\ & \Rightarrow a=\dfrac{-1}{6} \\\ \end{aligned}
So, substituting a=16a=\dfrac{-1}{6} in equation (i) and (ii) and solving for b and c, we get,
b=73b=\dfrac{-7}{3} and c=256c=\dfrac{-25}{6}
Now, substituting the values of a, b and c in the equation of parabola, we have,
y=x267x3256y=\dfrac{-{{x}^{2}}}{6}-\dfrac{7x}{3}-\dfrac{25}{6}
Now, to draw the graph, we have to determine the points at which the parabola will cut the x and y-axis.
So, for the parabola to cut the y-axis, x must be 0.
y=256\Rightarrow y=\dfrac{-25}{6}
And for parabola to cut x-axis, y must be 0.
0=x267x3256 x2+14x+25=0 x=14±14225×42 x=14±962 x=14±462 x=7+26 or 726 \begin{aligned} & \Rightarrow 0=\dfrac{-{{x}^{2}}}{6}-\dfrac{7x}{3}-\dfrac{25}{6} \\\ & \Rightarrow {{x}^{2}}+14x+25=0 \\\ & \Rightarrow x=\dfrac{-14\pm \sqrt{{{14}^{2}}-25\times 4}}{2} \\\ & \Rightarrow x=\dfrac{-14\pm \sqrt{96}}{2} \\\ & \Rightarrow x=\dfrac{-14\pm 4\sqrt{6}}{2} \\\ & \therefore x=-7+2\sqrt{6}\text{ }or\text{ }-7-2\sqrt{6} \\\ \end{aligned}
So, the graph can be plotted as shown below.

Note: One may note that there is an easy formula to solve the above question. The equation of parabola in vertex form is given as : y=a(xh)2+ky=a{{\left( x-h \right)}^{2}}+k, where (h,k)\left( h,k \right) is the vertex (7,4)\left( -7,4 \right). You may note an important thing that it is given that the axis of symmetry is x=7x=-7 and that is why we have assumed the equation of parabola as y=ax2+bx+cy=a{{x}^{2}}+bx+c. If the axis of symmetry would have been y = constant, then we would have assumed the equation of parabola as x=ay2+by+cx=a{{y}^{2}}+by+c.