Find equations for two lines through the origin that are tangent to the ellipse

7x2−168x+y2+62=0
please show me the steps :)

thanks :)

Why did the line bring a map to the ellipse's party? Because it wanted to find the perfect tangent!

To find the equations for two lines through the origin that are tangent to the given ellipse, we can follow these steps:

Step 1: Write the equation of the ellipse in standard form
Step 2: Find the derivative of the ellipse equation
Step 3: Find the slope of the tangent lines to the ellipse
Step 4: Use the slopes to find the equations of the lines through the origin

Let's go through each step in detail:

Step 1: Write the equation of the ellipse in standard form
The given equation of the ellipse is 7x^2 - 168x + y^2 + 62 = 0. To put it in standard form, we need to complete the square for both the x and y terms.

First, let's focus on the x terms:
7x^2 - 168x = -y^2 - 62
Divide the equation by 7 to simplify it:
x^2 - 24x = -y^2/7 -62/7

To complete the square, take half of the coefficient of x (which is -24/2 = -12), square it (which gives 144), and add it to both sides of the equation:
x^2 - 24x + 144 = -y^2/7 -62/7 + 144

Now, let's focus on the y terms:
-y^2/7 + 144 = -62/7
Multiply both sides of the equation by -7 to eliminate the fraction:
y^2 - 1008 = 62

Now, put together the x terms and y terms:
x^2 - 24x + 144 + y^2 - 1008 = 62
Simplify:
x^2 - 24x + y^2 - 946 = 0

Therefore, the equation of the ellipse in standard form is:
x^2 + y^2 - 24x - 946 = 0

Step 2: Find the derivative of the ellipse equation
To find the slope of the tangent lines to the ellipse, we need to differentiate the equation of the ellipse with respect to x.

Differentiating both sides of the equation:
d/dx (x^2 + y^2 - 24x - 946) = d/dx (0)
2x + 2yy' - 24 - 0 = 0
Simplify:
x + yy' - 12 = 0

Step 3: Find the slope of the tangent lines
We need to find the value of y' (the derivative of y with respect to x) when the tangent line passes through the origin (0, 0). Since the lines pass through the origin, the x value is 0.

Plug x = 0 into the equation we found in step 2:
0 + 0y' - 12 = 0
y' = 12

So, the slope of the tangent lines to the ellipse is 12.

Step 4: Use the slopes to find the equations of the lines through the origin
The tangent lines through the origin will have the form y = mx, where m is the slope.

For the first line, we have m = 12. Therefore, the equation of the first tangent line is:
y = 12x

For the second line, we know that the slope is the negative reciprocal of the first line's slope. So, the slope is -1/12. The equation of the second tangent line is:
y = (-1/12)x

Therefore, the equations for the two lines through the origin that are tangent to the ellipse are: y = 12x and y = (-1/12)x.

Rewrite the ellipse as:

7(x-12)²+y²-946=0
So it is an ellipse on the x-axis.

Check that the origin is exterior to the ellipse: set y=0,
x-12=±sqrt(946/7)=±11.625
x=0.375 or 23.625
=> the origin is exterior to the ellipse.

Slope of a point on the ellipse can be found by implicit differentiation of the ellipse equation:
14x-168+2y(dy/dx)=0 from which
dy/dx = (168-14x)/2y
Since the tangent line passes through the orgin, at the point of tangency, dy/dx=y/x, or
(168-14x)/2y = y/x
Transpose and solve for y:
y^2=84x-7x^2
Substitute into the ellipse equation to get y:
7x^2-168x+62+84x-7x^2=0
Solve for x:
84x=62 => x=31/42
Calculate y:
y=sqrt(84x-7x^2)
=sqrt(84*31/42-7(31/42)^2)
=sqrt(14663)/(6*sqrt(7))
=7.628 approx.

Therefore the tangent line is given by:
y=[sqrt(14663)/(6*sqrt(7))]/(31/42)x
=[sqrt(7)*sqrt(14663)/31]x

Check by graphing and check that the point is on the line.