# Find the value a and b such that the cone ax^2 + by^2 + z^2 = 0 and the ellips 2x^2 + 4y^2 + y^2 = 42 intersect and are perpendicular to each other at the point (1, 3, 2)

Find the value a and b such that the cone intersect and are perpendicular to each other at the point (1, 3, 2)

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Benedict

Step 1 We are given that the cone intersect and are perpendicular to each other at the point (1, 3, 2). So, both the equation intersect. Thus, $a{x}^{2}+b{y}^{2}+{z}^{2}=2{x}^{2}+4{y}^{2}+{z}^{2}-42$
$⇒a{x}^{2}-2{x}^{2}+b{y}^{2}-4{y}^{2}+{z}^{2}-{z}^{2}+42=0$
$⇒\left(a-2\right){x}^{2}+\left(b-4\right){y}^{2}+42=0$ ........(1) So, the locus of intersection of the cone and the ellipsoid is the circle written above. We further are given that both the conic sections are perpendicular to each other at the point (1, 3, 2). So, the equation (1) must be passing through the point (1, 3). Thus, $\left(a-2\right){x}^{2}+\left(b-4\right){y}^{2}+42=0$ $⇒\left(a-2\right)+4\left(b-4\right)+42=0$ $⇒a+4b+24=0$ ........(2) Step 3 Direction ratio of a function $F\left(x,y,z\right)=0$ is nothing but the gradient of F. So, the direction ratios of $a{x}^{2}+b{y}^{2}+{z}^{2}=0is<2ax,2by,2z>$. Then the direction ratios at $\left(1,3,2\right)is<2a,6b,4>$. Now, the direction ratios of $2{x}^{2}+4{y}^{2}+{z}^{2}-42=0is<4x,8y,2z>$. Then the direction ratios at $\left(1,3,2\right)is<4,24,4>$. We have that the conic sections are perpendicular to each other at the point (1, 3, 2). So, $8a+144b+16=0⇒a+18b+2=0$ .......(3) Step 4 Now, solving equation (2) and (3) to find the value of a and b. $a+18b+2=0$
$a+4b+24=0$ Isolate a for $a+18b+2=0:a=-18b-2$ Subtitute $a=-18b-2$ $-18b-2+4b+24=0$ Isolate $bfor-18b-2+4b+24=0:b=\frac{11}{7}$ For $a=-18b-2$ Subsititute $b=\frac{11}{7}$ The solution to the system of equations are: $b=\frac{11}{7},a=\frac{212}{7}$ Step 5 Thus, the equation of the cone is, $-\frac{212}{7}{x}^{2}+\frac{11}{7}{y}^{2}+{z}^{2}=0$