Find the equation of the tangent line
5x^2+y^2=14 at (1,3).
P.S: It's calculus.
So far I know this much
10x+2y(dy/dx)=o
-10x -10x
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2y(dy/dx)= -10x
(dy/dx)= -10x/2y= -5x/y
m= -5(1)/3= -5/3
y=mx+b
3= -5/3(1)+b then what?
ahh, so close!
solve for b in your last equation
I get b = 14/3
put m and b into your y = mx + b and you are done
To find the equation of the tangent line, you have correctly identified that the equation of the line is given by y = mx + b, where m is the slope of the line and b is the y-intercept.
You have already found the slope, m, which is -5/3. To determine the y-intercept, b, you need to substitute the coordinates (1,3) into the equation and solve for b.
Using the coordinates (1,3), you have the equation:
3 = (-5/3)(1) + b
To solve for b, you can simplify the equation:
3 = -5/3 + b
Next, you can add 5/3 to both sides of the equation:
3 + 5/3 = b
To simplify, you can find a common denominator for the fractions:
9/3 + 5/3 = b
14/3 = b
Now that you have found the value of b, you can substitute it back into the equation y = mx + b to get the equation of the tangent line:
y = (-5/3)x + 14/3
Therefore, the equation of the tangent line to the curve 5x^2 + y^2 = 14 at the point (1,3) is y = (-5/3)x + 14/3.