To solve this: differentiate both equations wrt x. Write m for the differentiation of y wrt x. Solve for m, this is the slope of the lines. Solve for pairs of (X,Y) These are the points on the equation where the slope is right. The equations of the tangent lines are: (y-Y) = m (x-X) where you substitute your found values for X,Y and m

You want to find tangent lines that are parallel to the line 9x-8y=1. Any line is parallel to another line if both lines have the same slope. Therefore, you must find the slope of the the given line. To find the slope of the given line use algebra to re-write it so that is in the form: y = mx + b I will leave the algebra to you but an important point in the remainder of my instructions to you is that the slope is positive. Next you need to take a look at the equation: x^2+16y^2=52 Because you are looking for tangent lines in the x-y plane, you have excluded the use of complex numbers so this equation is an eccentric elipse in the x-y plane. To give you an idea of the shape, let's look at the points where it intersects the x and y axis. To find the two points where it intercepts the x axis you make y = 0. This gives you the equation x^2 = 52 x = ±sqrt(52) The sqrt(52) is approximately 7.211 This means that the range for x in the x-y plane is: -7.211 ≤ x ≤ 7.211 Now we will do the y intercept by forcing x = 0: 16y^2 = 52 y^2 = 52/16 y = ±sqrt(52/16) The sqrt(52/16) is approximately 1.8 This means that the range for y in the x-y plane is: -1.8 ≤ y ≤ 1.8 If you take a moment to imagine this elipse, you can see that the tangent lines have a positive slope in the 2nd and 4th quandrants. Therefore, when the x value is positive, the corresponding value for y must be negative and, when the value for x is negative, the corresponding value for y must be positive. This is important, because we are about to take the derivative of the equation of the elipse and this will give use two different equations for x at the slope of the line. To take the derivative of x^2+16y^2=52 we first need to do some algebra: x^2 + 16y^2 = 52 16y^2 = 52 - x^2 y^2 = (1/16)(52 - x^2) y = ±(1/4){(52 - x^2)^(1/2)} dy/dx = ±(1/4)(-2x)(1/2){(52 - x^2)^(-1/2)} dy/dx = ±(x/4){(52 - x^2)^(-1/2)} For a positive slope the above equation becomes: m = (x/4){(52 - x^2)^(-1/2)} for x > 0 and m = -(x/4){(52 - x^2)^(-1/2)} for x < 0 When you solve these two equations you, will get a total of 4 values of x (both a positive value and a negative value for each equation) but choose only the positive value for the first and only the negative value for the second. You now have the two x values for two points on the elipse where the tangent has the same slope as the give line; let's call them x1 and x2. To find the corresponding y values substitute the value of x1 into the equation: x1^2+16y^2=52 You will get two values for y but you must remember to choose only the negative value; let's call that y1. Repeat this for x2 x2^2+16y^2=52 Again, you will get two values for y but you must remember to choose only the positive value for y; let's call that y2. Now you can find the equations of the two lines that have the same slope of the given line by substituting into the equation: y1 = m(x1) + b1 and solve for b1 y2 = m(x2) + b2 and solve for b2 Your two equations will be: y = mx + b1 and y = mx + b2 I hope that this helps.